Question

Find the intervals in which the following functions are strictly increasing or decreasing:
(a) $x^2+2x-5$
(b) $10-6x-2x^2$
(c) $-2x^3-9x^2-12x+1$
(d) $6-9x-x^2$
(e) $f\left(x\right)=(x+1{)}^3(x-3{)}^3$

Answer 1

(a)

We have,
$f\left(x\right)=x^2+2x-5$
$\therefore f^{\prime }\left(x\right)=2x+2$
Now, $f^{\prime }\left(x\right)=0\Rightarrow x=-1$
The point $x=-1$ divides the real line into two disjoint intervals
i.e., $(-\infty ,-1)$ and $(-1,\infty )$.
In interval $(-\infty ,-1),f^{\prime }\left(x\right)=2x+2<0$
$\therefore$ $f$ is strictly decreasing in interval $(-\infty ,-1)$.
And in interval $(-1,\infty ),f^{\prime }\left(x\right)=2x+2>0$
Thus, $f$ is strictly increasing for $x>1.$

(b)

$f\left(x\right)=10-6x-2x^2$
$\therefore f^{\prime }\left(x\right)=-6-4x$
Now, $f^{\prime }\left(x\right)=0\Rightarrow x=-\frac{3}{2}$
The point $x=-\frac{3}{2}$ divides the real line into two disjoint intervals i.e., $(-\infty ,-\frac{3}{2})$ and $(-\frac{3}{2},\infty )$.
In interval $(-\infty ,-\frac{3}{2})$ i.e., when $x<-\frac{3}{2},f^{\prime }\left(x\right)=-6-4x<0.$
$\therefore$ $f$ is strictly decreasing for $x<-\frac{3}{2}$ and in interval $(-\frac{3}{2},\infty ),f^{\prime }\left(x\right)>0$
Thus $f$ is strictly increasing in this interval.

(c)

$f\left(x\right)=-2x^3-9x^2-12x+1$
$\therefore f^{\prime }\left(x\right)=-6x^3-18x^2-12=-6(x^2+3x+2)=-6(x+1)(x+2)$
Now,
$f^{\prime }\left(x\right)=0\Rightarrow x=-1$ and $x=-2$
Points $x=-1$ and $x=-2$ divide the real line into three disjoint intervals
i.e., $(-\infty ,-2)(-2,-1)$ and $(-1,-\infty )$.
In intervals $(-\infty ,-2)$ and $(1,-\infty )$ i.e., when $x<-2$ and $x>-1,$
$f^{\prime }\left(x\right)=-6(x+1)(x+2)<0$
$\therefore$ $f$ is strictly decreasing for $x<-2$ and $x>-1$.
Now, in interval $(-2,-1)$ i.e., when $-2<x<-1,f^{\prime }\left(x\right)=-6(x+1)(x+2)>0$
$\therefore$ $f$ is strictly increasing for $-2<x<-1.$

(d)

$f\left(x\right)=6-9x-x^2$
$f^{\prime }\left(x\right)=-9-2x$

For f to be strictly increasing $f^{\prime }\left(x\right)>0\Rightarrow 2x+9<0\Rightarrow x<-\frac{9}{2}$
And for f to be strictly decreasing, $f^{\prime }\left(x\right)<0\Rightarrow 2x+9>0\Rightarrow x>-\frac{9}{2}$

(e)

We have,
$f\left(x\right)=(x+1{)}^3(x-3{)}^3$
$f^{\prime }\left(x\right)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$
$=3(x+1{)}^2(x-3{)}^2[x-3+x+1]$
$=3(x+1{)}^2(x-3{)}^2(2x-2)$
$=6(x+1{)}^2(x-3{)}^2(x-1)$
Now,
$f^{\prime }\left(x\right)=0\Rightarrow x=-1,3,1$
The points $x=-1,x=1,$ and $x=3$ divide the real line into four disjoint intervals
i.e.,$(-\infty ,-1),(-1,1),(1,3)$ and $(3,\infty )$.
In intervals $(-\infty ,-1)$ and $(-1,1)$, $f^{\prime }\left(x\right)=6(x+1{)}^2(x-3{)}^2(x-1)<0$
$\therefore$ $f$ is strictly decreasing in intervals $(-\infty ,-1)$ and $(-1,1).$
In intervals $(1,3)$ and $(3,\infty )f^{\prime }\left(x\right)=6(x+1{)}^2(x-3{)}^2(x-1)>0$
$\therefore$ $f$ is strictly increasing in intervals $(1,3)$ and $(3,\infty )$.