Question

Find the absolue maximum value and he absolute minimum value of the following functions in the given intervals:

$f\left(x\right)=x^3,xϵ[-2,2]$

$f\left(x\right)=sinx+cosx,xϵ[0,\pi ]$

$f\left(x\right)=4x-\frac{1}{2}{x}^2,xϵ[-2,\frac{9}{2}]$

$f\left(x\right)=(x-1{)}^2+3,xϵ[-3,1]$

Answer 1

Given function is, $f\left(x\right)=x^3,\Rightarrow f^{\prime }\left(x\right)=3x^2$
For maxima or minima put $f^{\prime }\left(x\right)=0\Rightarrow 3x^2=0\Rightarrow x=0ϵ[-2,2]$
Now, we evaluate the value of f at critical point x=0 and at end points of the interval [-2,2].
$\begin{array}{cc}Atx=0& f\left(0\right)=0^3=0\\ Atx=-2& f(-2)=(-2{)}^3=-8\\ ATx=2& f\left(2\right)=(2{)}^3=8\end{array}$
Thus, absolute maximum value is 8 at x=2 and absolute minimum value is -8 at x=-2

Given function $f\left(x\right)=sinx+cosx\Rightarrow f^{\prime }\left(x\right)=cosx-sinx$
For maxima or minima put $f^{\prime }\left(x\right)=0\Rightarrow cos-sinx=0\Rightarrow \frac{sinx}{cosx}=1$
$\Rightarrow tanx=1\Rightarrow \frac{\pi }{4}ϵ[0,\pi ]$
Now, we evaluate the value of f at critical point $x=\frac{\pi }{4}$ and at the end points of the interval $[0,\pi ]$.
$\begin{array}{cc}Atx=\frac{\pi }{4}& f\left(\frac{\pi }{4}\right)=sin\frac{\pi }{4}+cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\\ Atx=0,& f\left(0\right)=sin0+cos0=0+1=1\\ Atx=\pi & f\left(\pi \right)=sin\pi +cos\pi =0-1=-1\end{array}$
Thus, absolute maximum value is $\sqrt{2}atx=\frac{\pi }{4}$ and absolute minimum value is -1 at $x=\pi$

Given function is, $f\left(x\right)=4x-\frac{1}{2}{x}^2,\Rightarrow f^{\prime }\left(x\right)=4-\frac{1}{2}\left(2x\right)=4-x$
For maxima or minima put $f^{\prime }\left(x\right)=0,4-x=0\Rightarrow x=4ϵ[-2,\frac{9}{2}]$
Now, we evaluate the value of f at critical point x=4 and at the end points of the interval $[-2,\frac{9}{2}]$
$\begin{array}{cc}Atx=4& f\left(4\right)=4\left(4\right)-\frac{1}{2}(4{)}^2=16-8=8\\ Atx=-2& f(-2)=4(-2)-\frac{1}{2}(-2{)}^2=-8-2=-10\\ Atx=\frac{9}{2},& f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}{\left(\frac{9}{2}\right)}^2=18-\frac{81}{8}=\frac{63}{8}=7.875\end{array}$
Thus, absolute maximum value is 8 at x=4 and absolute minimum value is -10 at x=-2.

Given function is, $f\left(x\right)=(x-1{)}^2+3,$
$\therefore f^{\prime }\left(x\right)=2(x-1)$
For maxima or minima put $f^{\prime }\left(x\right)=0\Rightarrow 2(x-1)=0\Rightarrow x=1$
Now, we evaluate the value of f at critical point x=1 and at the end points of the interval [-3,1].
$\begin{array}{cc}Atx=1& f\left(1\right)=(1-1{)}^2+3=3\\ ATx=-3& f(-3)=(-3-1{)}^2+3=19\end{array}$
Thus, absolute maximum value is 19 at x=-3 and absolute minimum value is 3 at x=1