Question

Find the intervals in which the following functions are increasing or decreasing.
(i) f(x) = 10 − 6x − 2x²

(ii) f(x) = x² + 2x − 5

(iii) f(x) = 6 − 9x − x²

(iv) f(x) = 2x³ − 12x² + 18x + 15

(v) f(x) = 5 + 36x + 3x² − 2x³

(vi) f(x) = 8 + 36x + 3x² − 2x³

(vii) f(x) = 5x³ − 15x² − 120x + 3

(viii) f(x) = x³ − 6x² − 36x + 2

(ix) f(x) = 2x³ − 15x² + 36x + 1

(x) f(x) = 2x³ + 9x² + 12x + 20

(xi) f(x) = 2x³ − 9x² + 12x − 5

(xii) f(x) = 6 + 12x + 3x² − 2x³

(xiii) f(x) = 2x³ − 24x + 107

(xiv) f(x) = −2x³ − 9x² − 12x + 1

(xv) f(x) = (x − 1) (x − 2)²

(xvi) f(x) = x³ − 12x² + 36x + 17

(xvii) f(x) = 2x³ − 24x + 7

(xviii) $f\left(x\right)=\frac{3}{10}{x}^4-\frac{4}{5}{x}^3-3x^2+\frac{36}{5}x+11$

(xix) f(x) = x⁴ − 4x

(xx) $f\left(x\right)=\frac{x^4}{4}+\frac{2}{3}{x}^3-\frac{5}{2}{x}^2-6x+7$

(xxi) f(x) = x⁴ − 4x³ + 4x² + 15

(xxii) f(x) = 5x^3/2 − 3x^5/2, x > 0

(xxiii) f(x) = x⁸ + 6x²

(xxiv) f(x) = x³ − 6x² + 9x + 15

(xxv) $f\left(x\right)={\left\{x(x-2)\right\}}^2$

Answer 1

$\mathrm{When}\left(x-a\right)\left(x-b\right)\text{>0 with}a<b,x<a\text{or}\mathit{\text{x>b.}}$

$\mathrm{When}\left(x-a\right)\left(x-b\right)\text{<0 with}a<b,a<x<b.$

$$\begin{gathered} \left(\mathrm{i}\right) \\ f\left(x\right)=10-6x-2x^2 \\ f\text{'}\left(x\right)=-6-4x \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{increasing},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow -6-4x>0 \\ \Rightarrow -4x>6 \\ \Rightarrow x<\frac{-3}{2} \\ \Rightarrow x\in \left(-\infty ,\frac{-3}{2}\right) \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{on}\left(-\infty ,\frac{-3}{2}\right). \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{decreasing},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow -6-4x<0 \\ \Rightarrow -4x<6 \\ \Rightarrow x>\frac{-6}{4} \\ \Rightarrow x>\frac{-3}{2} \\ \Rightarrow x\in \left(\frac{-3}{2},\infty \right) \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{decreasing}\mathrm{on}\left(\frac{-3}{2},\infty \right). \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ f\left(x\right)=x^2+2x-5 \\ f\text{'}\left(x\right)=2x+2 \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 2x+2>0 \\ \Rightarrow 2\left(x+1\right)>0 \\ \Rightarrow x+1>0 \\ \Rightarrow x>-1 \\ \Rightarrow x\in \left(-1,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(-1,\infty \right). \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 2x+2<0 \\ \Rightarrow 2\left(x+1\right)<0 \\ \Rightarrow x+1<0 \\ \Rightarrow x<-1 \\ \Rightarrow x\in \left(-\infty ,-1\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-\infty ,-1\right)\text{.} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \\ f\left(x\right)=6-9x-x^2 \\ f\text{'}\left(x\right)=-2x-9 \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow -2x-9>0 \\ \Rightarrow -2x>9 \\ \Rightarrow x<\frac{-9}{2} \\ \Rightarrow x\in \left(-\infty ,\frac{-9}{2}\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(-\infty ,\frac{-9}{2}\right). \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow -2x-9<0 \\ \Rightarrow -2x<9 \\ \Rightarrow x>\frac{-9}{2} \\ \Rightarrow x\in \left(\frac{-9}{2},\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(\frac{-9}{2},\infty \right)\text{.} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \\ f\left(x\right)=2x^3-12x^2+18x+15 \\ f\text{'}\left(x\right)=6x^2-24x+18 \\ =6\left(x^2-4x+3\right) \\ =6\left(x-1\right)\left(x-3\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 6\left(x-1\right)\left(x-3\right)>0 \\ \Rightarrow \left(x-1\right)\left(x-3\right)>0\left[\mathrm{Since}6>0,6\left(x-1\right)\left(x-3\right)>0\Rightarrow \left(x-1\right)\left(x-3\right)>0\right] \\ \Rightarrow x<1\text{or}x>3 \\ \Rightarrow x\in \left(-\infty ,1\right)\cup \left(3,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(-\infty ,1\right)\cup \left(3,\infty \right). \end{gathered}$$




$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 6\left(x-1\right)\left(x-3\right)<0 \\ \Rightarrow \left(x-1\right)\left(x-3\right)<0\left[\mathrm{Since}6>0,6\left(x-1\right)\left(x-3\right)<0\Rightarrow \left(x-1\right)\left(x-3\right)<0\right] \\ \Rightarrow 1<x<3 \\ \Rightarrow x\in \left(1,3\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(1,3\right)\text{.} \end{gathered}$$




$$\begin{gathered} \left(\mathrm{v}\right) \\ f\left(x\right)=5+36x+3x^2-2x^3 \\ f\text{'}\left(x\right)=36+6x-6x^2 \\ =-6\left(x^2-x-6\right) \\ =-6\left(x-3\right)\left(x+2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow -6\left(x-3\right)\left(x+2\right)>0 \\ \Rightarrow \left(x-3\right)\left(x+2\right)<0\left[\mathrm{Since}-6<0,-6\left(x-1\right)\left(x+2\right)>0\Rightarrow \left(x-1\right)\left(x+2\right)<0\right] \\ \Rightarrow -2<x<3 \\ \Rightarrow x\in \left(-2,3\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(-2,3\right). \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow -6\left(x-3\right)\left(x+2\right)<0 \\ \Rightarrow \left(x-3\right)\left(x+2\right)>0\left[\mathrm{Since}-6<0,-6\left(x-1\right)\left(x+2\right)<0\Rightarrow \left(x-1\right)\left(x+2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>3 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(3,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-\infty ,-2\right)\cup \left(3,\infty \right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vi}\right) \\ f\left(x\right)=8+36x+3x^2-2x^3 \\ f\text{'}\left(x\right)=36+6x-6x^2 \\ =-6\left(x^2-x-6\right) \\ =-6\left(x-3\right)\left(x+2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow -6\left(x-3\right)\left(x+2\right)>0 \\ \Rightarrow \left(x-3\right)\left(x+2\right)<0\left[\mathrm{Since}-6<0,-6\left(x-3\right)\left(x+2\right)>0\Rightarrow \left(x-3\right)\left(x+2\right)<0\right] \\ \Rightarrow -2<x<3 \\ \Rightarrow x\in \left(-2,3\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(-2,3\right). \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow -6\left(x-3\right)\left(x+2\right)<0 \\ \Rightarrow \left(x-3\right)\left(x+2\right)>0\left[\mathrm{Since}-6<0,-6\left(x-3\right)\left(x+2\right)<0\Rightarrow \left(x-3\right)\left(x+2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>3 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(3,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-\infty ,-2\right)\cup \left(3,\infty \right). \end{gathered}$$



$$\begin{gathered} \left(\mathrm{vii}\right) \\ f\left(x\right)=5x^3-15x^2-120x+3 \\ f\text{'}\left(x\right)=15x^2-30x-120 \\ =15\left(x^2-2x-8\right) \\ =15\left(x-4\right)\left(x+2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 15\left(x-4\right)\left(x+2\right)>0 \\ \Rightarrow \left(x-4\right)\left(x+2\right)>0\left[\mathrm{Since}15>0,15\left(x-4\right)\left(x+2\right)>0\Rightarrow \left(x-4\right)\left(x+2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>4 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(4,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,-2\right)\cup \left(4,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have,} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 15\left(x-4\right)\left(x+2\right)<0 \\ \Rightarrow \left(x-4\right)\left(x+2\right)<0\left[\mathrm{Since}15>0,15\left(x-4\right)\left(x+2\right)<0\Rightarrow \left(x-4\right)\left(x+2\right)<0\right] \\ \Rightarrow -2<x<4 \\ \Rightarrow x\in \left(-2,4\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-2,4\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{viii}\right) \\ f\left(x\right)=x^3-6x^2-36x+2 \\ f\text{'}\left(x\right)=3x^2-12x-36 \\ =3\left(x^2-4x-12\right) \\ =3\left(x-6\right)\left(x+2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have,} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 3\left(x-6\right)\left(x+2\right)>0 \\ \Rightarrow \left(x-6\right)\left(x+2\right)>0\left[\mathrm{Since}3>0,3\left(x-6\right)\left(x+2\right)>0\Rightarrow \left(x-6\right)\left(x+2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>6 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(6,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,-2\right)\cup \left(6,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 3\left(x-6\right)\left(x+2\right)<0 \\ \Rightarrow \left(x-6\right)\left(x+2\right)<0\left[\mathrm{Since}3>0,3\left(x-6\right)\left(x+2\right)<0\Rightarrow \left(x-6\right)\left(x+2\right)<0\right] \\ \Rightarrow -2<x<6 \\ \Rightarrow x\in \left(-2,6\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-2,6\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ix}\right) \\ f\left(x\right)=2x^3-15x^2+36x+1 \\ f\text{'}\left(x\right)=6x^2-30x+36 \\ =6\left(x^2-5x+6\right) \\ =6\left(x-2\right)\left(x-3\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 6\left(x-2\right)\left(x-3\right)>0 \\ \Rightarrow \left(x-2\right)\left(x-3\right)>0\left[\mathrm{Since}6>0,6\left(x-2\right)\left(x-3\right)>0\Rightarrow \left(x-2\right)\left(x-3\right)>0\right] \\ \Rightarrow x<2\text{or}x>3 \\ \Rightarrow x\in \left(-\infty ,2\right)\cup \left(3,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,2\right)\cup \left(3,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 6\left(x-2\right)\left(x-3\right)<0 \\ \Rightarrow \left(x-2\right)\left(x-3\right)<0\left[\mathrm{Since}6>0,6\left(x-2\right)\left(x-3\right)<0\Rightarrow \left(x-2\right)\left(x-3\right)<0\right] \\ \Rightarrow 2<x<3 \\ \Rightarrow x\in \left(2,3\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(2,3\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{x}\right) \\ f\left(x\right)=2x^3+9x^2+12x+20 \\ f\text{'}\left(x\right)=6x^2+18x+12 \\ =6\left(x^2+3x+2\right) \\ =6\left(x+1\right)\left(x+2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 6\left(x+1\right)\left(x+2\right)>0 \\ \Rightarrow \left(x+1\right)\left(x+2\right)>0\left[\mathrm{Since}6>0,6\left(x+1\right)\left(x+2\right)>0\Rightarrow \left(x+1\right)\left(x+2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>-1 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(-1,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,-2\right)\cup \left(-1,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 6\left(x+1\right)\left(x+2\right)<0 \\ \Rightarrow \left(x+1\right)\left(x+2\right)<0\left[\mathrm{Since}6>0,6\left(x+1\right)\left(x+2\right)<0\Rightarrow \left(x+1\right)\left(x+2\right)<0\right] \\ \Rightarrow -2<x<-1 \\ \Rightarrow x\in \left(-2,-1\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-2,-1\right). \end{gathered}$$



$$\begin{gathered} \left(\mathrm{xi}\right) \\ f\left(x\right)=2x^3-9x^2+12x-5 \\ f\text{'}\left(x\right)=6x^2-18x+12 \\ =6\left(x^2-3x+2\right) \\ =6\left(x-1\right)\left(x-2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 6\left(x-1\right)\left(x-2\right)>0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)>0\left[\mathrm{Since}6>0,6\left(x-1\right)\left(x-2\right)>0\Rightarrow \left(x-1\right)\left(x-2\right)>0\right] \\ \Rightarrow x<1\text{or}x>2 \\ \Rightarrow x\in \left(-\infty ,1\right)\cup \left(2,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 6\left(x-1\right)\left(x-2\right)<0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)<0\left[\mathrm{Since}6>0,6\left(x-1\right)\left(x-2\right)<0\Rightarrow \left(x-1\right)\left(x-2\right)<0\right] \\ \Rightarrow 1<x<2 \\ \Rightarrow x\in \left(1,2\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(1,2\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xii}\right) \\ f\left(x\right)=6+12x+3x^2-2x^3 \\ f\text{'}\left(x\right)=12+6x-6x^2 \\ =-6\left(x^2-x-2\right) \\ =-6\left(x-2\right)\left(x+1\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow -6\left(x-2\right)\left(x+1\right)>0 \\ \Rightarrow \left(x-2\right)\left(x+1\right)<0\left[\mathrm{Since}-6<0,-6\left(x-2\right)\left(x+1\right)>0\Rightarrow \left(x-2\right)\left(x+1\right)<0\right] \\ \Rightarrow -1<x<2 \\ \Rightarrow x\in \left(-1,2\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(-1,2\right). \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow -6\left(x-2\right)\left(x+1\right)<0 \\ \Rightarrow \left(x-2\right)\left(x+1\right)>0\left[\mathrm{Since}-6<0,-6\left(x-2\right)\left(x+1\right)<0\Rightarrow \left(x-2\right)\left(x+1\right)>0\right] \\ \Rightarrow x<-1\text{or}x>2 \\ \Rightarrow x\in \left(-\infty ,-1\right)\cup \left(2,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-\infty ,-1\right)\cup \left(2,\infty \right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xiii}\right) \\ f\left(x\right)=2x^3-24x+107 \\ f\text{'}\left(x\right)=6x^2-24=6\left(x^2-4\right)=6\left(x+2\right)\left(x-2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 6\left(x+2\right)\left(x-2\right)>0 \\ \Rightarrow \left(x+2\right)\left(x-2\right)>0\left[\mathrm{Since}6>0,6\left(x+2\right)\left(x-2\right)>0\Rightarrow \left(x+2\right)\left(x-2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>2 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(2,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 6\left(x+2\right)\left(x-2\right)<0 \\ \Rightarrow \left(x+2\right)\left(x-2\right)<0\left[\mathrm{Since}6>0,6\left(x+2\right)\left(x-2\right)<0\Rightarrow \left(x+2\right)\left(x-2\right)<0\right] \\ \Rightarrow -2<x<2 \\ \Rightarrow x\in \left(-2,2\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-2,2\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xiv}\right) \\ f\left(x\right)=-2x^3-9x^2-12x+1 \\ f\text{'}\left(x\right)=-6x^2-18x-12 \\ =-6\left(x^2+3x+2\right) \\ =-6\left(x+1\right)\left(x+2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow -6\left(x+1\right)\left(x+2\right)>0 \\ \Rightarrow \left(x+1\right)\left(x+2\right)<0\left[\mathrm{Since}-6<0,-6\left(x+1\right)\left(x+2\right)>0\Rightarrow \left(x+1\right)\left(x+2\right)<0\right] \\ \Rightarrow -2<x<-1 \\ \Rightarrow x\in \left(-2,-1\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(-2,-1\right). \end{gathered}$$






$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow -6\left(x+1\right)\left(x+2\right)<0 \\ \Rightarrow \left(x+1\right)\left(x+2\right)>0\left[\mathrm{Since}-6<0,-6\left(x+1\right)\left(x+2\right)<0\Rightarrow \left(x+1\right)\left(x+2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>-1 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(-1,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-\infty ,-2\right)\cup \left(-1,\infty \right). \end{gathered}$$



$$\begin{gathered} \left(\mathrm{xv}\right) \\ f\left(x\right)=\left(x-1\right){\left(x-2\right)}^2 \\ =\left(x-1\right)\left(x^2-4x+4\right) \\ =x^3-5x^2+8x-4 \\ f\text{'}\left(x\right)=3x^2-10x+8 \\ =3x^2-6x-4x+8 \\ =\left(x-2\right)\left(3x-4\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \left(x-2\right)\left(3x-4\right)>0 \\ \Rightarrow x<\frac{4}{3}\text{or}x>2 \\ \Rightarrow x\in \left(-\infty ,-\frac{4}{3}\right)\cup \left(2,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,\frac{4}{3}\right)\cup \left(2,\infty \right)\text{.} \end{gathered}$$


$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow \left(x-2\right)\left(3x-4\right)<0 \\ \Rightarrow \frac{4}{3}<x<2 \\ \Rightarrow x\in \left(\frac{4}{3},2\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(\frac{4}{3},2\right). \end{gathered}$$



$$\begin{gathered} \left(\mathrm{xvi}\right) \\ f\left(x\right)=x^3-12x^2+36x+17 \\ f\text{'}\left(x\right)=3x^2-24x+36 \\ =3\left(x^2-8x+12\right) \\ =3\left(x-2\right)\left(x-6\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 3\left(x-2\right)\left(x-6\right)>0 \\ \Rightarrow \left(x-2\right)\left(x-6\right)>0\left[\mathrm{Since}3>0,3\left(x-2\right)\left(x-6\right)>0\Rightarrow \left(x-2\right)\left(x-6\right)>0\right] \\ \Rightarrow x<2\text{or}x>6 \\ \Rightarrow x\in \left(-\infty ,2\right)\cup \left(6,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,2\right)\cup \left(6,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 3\left(x-2\right)\left(x-6\right)<0 \\ \Rightarrow \left(x-2\right)\left(x-6\right)<0\left[\mathrm{Since}3>0,3\left(x-2\right)\left(x-6\right)<0\Rightarrow \left(x-2\right)\left(x-6\right)<0\right] \\ \Rightarrow 2<x<6 \\ \Rightarrow x\in \left(2,6\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(2,6\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xvii}\right) \\ f\left(x\right)=2x^3-24x+7 \\ f\text{'}\left(x\right)=6x^2-24 \\ =6\left(x^2-4\right) \\ =6\left(x+2\right)\left(x-2\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 6\left(x+2\right)\left(x-2\right)>0 \\ \Rightarrow \left(x+2\right)\left(x-2\right)>0\left[\mathrm{Since}6>0,6\left(x+2\right)\left(x-2\right)>0\Rightarrow \left(x+2\right)\left(x-2\right)>0\right] \\ \Rightarrow x<-2\text{or}x>2 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(2,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 6\left(x+2\right)\left(x-2\right)<0 \\ \Rightarrow \left(x+2\right)\left(x-2\right)<0\left[\mathrm{Since}6>0,6\left(x+2\right)\left(x-2\right)<0\Rightarrow \left(x+2\right)\left(x-2\right)<0\right] \\ \Rightarrow -2<x<2 \\ \Rightarrow x\in \left(-2,2\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-2,2\right). \end{gathered}$$



$$\begin{gathered} \left(\mathrm{xviii}\right) \\ f\left(x\right)=\frac{3}{10}{x}^4-\frac{4}{5}{x}^3-3x^2+\frac{36}{5}x+11 \\ =\frac{3x^4-8x^3-30x^2+72x+110}{10} \\ f\text{'}\left(x\right)=\frac{12x^3-24x^2-60x+72}{10} \\ =\frac{12}{10}\left(x^3-2x^2-5x+6\right) \\ =\frac{\left(x-1\right)\left(x^2-x-6\right)}{10} \\ =\frac{12}{10}\left(x-1\right)\left(x+2\right)\left(x-3\right) \\ \mathrm{Here},1,2\mathrm{and}3\mathrm{are}\mathrm{the}\mathrm{critical}\mathrm{points}. \\ \text{The possible intervals are}\left(-\infty -2\right)\text{,}\left(-2,1\right)\text{,}\left(1,3\right)\text{and}\left(3,\infty \right)\text{.} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \frac{12}{10}\left(x-1\right)\left(x+2\right)\left(x-3\right)>0 \\ \Rightarrow \left(x-1\right)\left(x+2\right)\left(x-3\right)>0 \\ \Rightarrow x\in \left(-2,1\right)\cup \left(3,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-2,1\right)\cup \left(3,\infty \right). \end{gathered}$$




$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow \frac{12}{10}\left(x-1\right)\left(x+2\right)\left(x-3\right)<0 \\ \Rightarrow \left(x-1\right)\left(x+2\right)\left(x-3\right)<0 \\ \Rightarrow x\in \left(-\infty -2\right)\cup \left(1,3\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-\infty -2\right)\cup \left(1,3\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xix}\right) \\ f\left(x\right)=x^4-4x \\ f\text{'}\left(x\right)=4x^3-4 \\ =4\left(x^3-1\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 4\left(x^3-1\right)>0 \\ \Rightarrow x^3-1>0 \\ \Rightarrow x^3>1 \\ \Rightarrow x>1 \\ \Rightarrow x\in \left(1,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(1,\infty \right). \end{gathered}$$


$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 4\left(x^3-1\right)<0 \\ \Rightarrow x^3-1<0 \\ \Rightarrow x^3<1 \\ \Rightarrow x<1 \\ \Rightarrow x\in \left(-\infty ,1\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-\infty ,1\right)\text{.} \end{gathered}$$



$$\begin{gathered} \left(\mathrm{xx}\right) \\ f\left(x\right)=\frac{x^4}{4}+\frac{2}{3}{x}^3-\frac{5}{2}{x}^2-6x+7 \\ =\frac{3x^4+8x^3-30x^2-72x+84}{12} \\ f\text{'}\left(x\right)=\frac{12x^3+24x^2-60x-72}{12} \\ =\left(x^3+2x^2-5x-6\right) \\ =\left(x+1\right)\left(x^2+x-6\right) \\ =\left(x+1\right)\left(x-2\right)\left(x+3\right) \\ \mathrm{Here},-1,\; 2\; and\; -3\mathrm{are}\mathrm{the}\mathrm{critical}\mathrm{points}\text{.} \\ \text{The possible intervals are}\left(\mathit{-}\mathit{\infty }\mathit{}\mathit{-}\mathit{3}\right)\mathit{\text{,}}\left(\mathit{-}\mathit{3}\mathit{,}\mathit{}\mathit{-}\mathit{1}\right)\mathit{\text{,}}\left(\mathit{-}\mathit{1}\mathit{,}\mathit{}\mathit{2}\right)\text{and}\left(2,\infty \right)\text{.} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \left(x+1\right)\left(x-2\right)\left(x+3\right)>0 \\ \Rightarrow x\in \left(-3,-1\right)\cup \left(2,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-3,-1\right)\cup \left(2,\infty \right). \end{gathered}$$



$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow \left(x+1\right)\left(x-2\right)\left(x+3\right)<0 \\ \Rightarrow x\in \left(-\infty -3\right)\cup \left(-1,2\right)\left[\text{From eq. (1)}\right] \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-\infty -3\right)\cup \left(-1,2\right). \end{gathered}$$






$$\begin{gathered} \left(\mathrm{xxi}\right) \\ f\left(x\right)=x^4-4x^3+4x^2+15 \\ f\text{'}\left(x\right)=4x^3-12x^2+8x \\ =4x\left(x^2-3x+2\right) \\ =4x\left(x-1\right)\left(x-2\right) \\ \mathrm{Here},\mathrm{0,\; 1\; and\; 2}\mathrm{are}\mathrm{the}\mathrm{c}\text{ritical points.} \\ \text{The possible intervals are}\left(-\infty ,0\right)\text{,}\left(0,1\right)\text{,}\left(1,2\right)\text{and}\left(2,\infty \right)\text{. ...(1)} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 4x\left(x-1\right)\left(x-2\right)>0\left[\mathrm{Since}4>0,4x\left(x-1\right)\left(x-2\right)>0\Rightarrow x\left(x-1\right)\left(x-2\right)>0\right] \\ \Rightarrow x\left(x-1\right)\left(x-2\right)>0 \\ \Rightarrow x\in \left(0,1\right)\cup \left(2,\infty \right)\left[\text{From eq. (1)}\right] \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(0,1\right)\cup \left(2,\infty \right). \end{gathered}$$



$$\begin{gathered} \mathrm{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 4x\left(x-1\right)\left(x-2\right)<0\left[\mathrm{Since}4>0,4x\left(x-1\right)\left(x-2\right)<0\Rightarrow x\left(x-1\right)\left(x-2\right)<0\right] \\ \Rightarrow x\left(x-1\right)\left(x-2\right)<0 \\ \Rightarrow x\in \left(-\infty ,0\right)\cup \left(1,2\right)\left[\text{From eq. (1)}\right] \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-\infty ,0\right)\cup \left(1,2\right). \end{gathered}$$





$$\begin{gathered} \left(\mathrm{xxii}\right)f\left(x\right)=5x^{\frac{3}{2}}-3x^{\frac{5}{2}},x>0 \\ f\text{'}\left(x\right)=\frac{15}{2}{x}^{\frac{1}{2}}-\frac{15}{2}{x}^{\frac{3}{2}} \\ =\frac{15}{2}{x}^{\frac{1}{2}}\left(1-x\right) \\ \mathrm{Here},0,1\mathrm{are}\mathrm{t}\text{he roots.} \\ \text{The possible intervals are}\left(-\infty ,0\right)\text{,}\left(0,1\right)\text{and}\left(1,\infty \right)\text{...(1)} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \frac{15}{2}{x}^{\frac{1}{2}}\left(1-x\right)>0 \\ \Rightarrow x\in \left(0,1\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(0,1\right). \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow \frac{15}{2}{x}^{\frac{1}{2}}\left(1-x\right)<0 \\ \Rightarrow x\in \left(1,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(1,\infty \right)\text{.} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xxiii}\right) \\ f\left(x\right)=x^8+6x^2 \\ f\text{'}\left(x\right)=8x^7+12x \\ =4x\left(2x^6+3\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 4x\left(2x^6+3\right)>0\left[\mathrm{Since}\left(2x^6+3\right)>0,4x\left(2x^6+3\right)>0\Rightarrow x>0\right] \\ \Rightarrow x>0 \\ \Rightarrow x\in \left(0,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(0,\infty \right). \end{gathered}$$



$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 4x\left(2x^6+3\right)<0 \\ \Rightarrow x<0\left[\mathrm{Since}\left(2x^6+3\right)>0,4x\left(2x^6+3\right)<0\Rightarrow x<0\right] \\ \Rightarrow x\in \left(-\infty ,0\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(-\infty ,0\right). \end{gathered}$$






$$\begin{gathered} \left(\mathrm{xxiv}\right) \\ f\left(x\right)=x^3-6x^2+9x+15 \\ f\text{'}\left(x\right)=3x^2-12x+9 \\ =3\left(x^2-4x+3\right) \\ =3\left(x-1\right)\left(x-3\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 3\left(x-1\right)\left(x-3\right)>0 \\ \Rightarrow \left(x-1\right)\left(x-3\right)>0\left[\mathrm{Since}3>0,3\left(x-1\right)\left(x-3\right)>0\Rightarrow \left(x-1\right)\left(x-3\right)>0\right] \\ \Rightarrow x<1\text{or}x>3 \\ \Rightarrow x\in \left(-\infty ,1\right)\cup \left(3,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(-\infty ,1\right)\cup \left(3,\infty \right)\text{.} \end{gathered}$$





$$\begin{gathered} \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 3\left(x-1\right)\left(x-3\right)<0 \\ \Rightarrow \left(x-1\right)\left(x-3\right)<0\left[\mathrm{Since}3>0,3\left(x-1\right)\left(x-3\right)<0\Rightarrow \left(x-1\right)\left(x-3\right)<0\right] \\ \Rightarrow 1<x<3 \\ \Rightarrow x\in \left(1,3\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}x\in \left(1,3\right). \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xxv}\right) \\ f\left(x\right)={\left\{x\left(x-2\right)\right\}}^2 \\ ={\left(x^2-2x\right)}^2 \\ =x^4+4x^2-4x^3 \\ f\text{'}\left(x\right)=4x^3+8x-12x^2 \\ =4x\left(x^2-3x+2\right) \\ =4x\left(x-1\right)\left(x-2\right) \\ \mathrm{Here},\mathrm{0,\; 1\; and\; 2}\mathrm{are}\mathrm{the}\mathrm{c}\text{ritical points.} \\ \text{The possible intervals are}\left(-\infty ,0\right)\text{,}\left(0,1\right)\text{,}\left(1,2\right)\text{and}\left(2,\infty \right)\text{.} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 4x\left(x-1\right)\left(x-2\right)>0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)>0 \\ \Rightarrow x\in \left(0,1\right)\cup \left(2,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}x\in \left(0,1\right)\cup \left(2,\infty \right). \end{gathered}$$





$$\begin{gathered} \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{decreasing},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 4x\left(x-1\right)\left(x-2\right)<0 \\ \Rightarrow x\left(x-1\right)\left(x-2\right)<0 \\ \Rightarrow x\in \left(-\infty ,0\right)\cup \left(1,2\right) \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{decreasing}\mathrm{on}x\in \left(-\infty ,0\right)\cup \left(1,2\right). \end{gathered}$$