Question

1)Find the maximum and minimum values, if any, of the function given by
$f\left(x\right)=|x+2|-1$

2)Find the maximum and minimum values, if any, of the function given by
$g\left(x\right)=-|x+1|+3$

3)Find the maximum and minimum values, if any, of the function given by
$h\left(x\right)=\sin \left(2x\right)+5$

4) Find the maximum and minimum values, if any, of the function given by
$f\left(x\right)=|\sin 4x+3|$

5) Find the maximum and minimum values, if any, of the function given by
$h\left(x\right)=x+1,xϵ(-1,1)$

Answer 1

1) $f\left(x\right)=|x+2|-1$

We know,

$|x+2|\ge 0$

$\Rightarrow |x+2|-1\ge -1$

$\Rightarrow f\left(x\right)\ge -1$

Hence, minimum value of $f\left(x\right)$ is $-1$ and there is no maximum value of $f\left(x\right)$.

2) $g\left(x\right)=-|x+1|+3$

We know,

$|x+1|\ge 0$

$\Rightarrow -|x+1|\le 0$

$\Rightarrow -|x+1|+3\le 3$

$\Rightarrow g\left(x\right)\le 3$

Hence, maximum value of $g\left(x\right)$ is $3$ and there is no minimum value of $g\left(x\right)$.

3) $h\left(x\right)=\sin \left(2x\right)+5$

We know,

$\sin \theta ϵ[-1,1]$

$\Rightarrow \sin 2xϵ[-1,1]$

$\Rightarrow -1\le \sin 2x\le 1$

$\Rightarrow -1+5\le \sin 2x+5\le 1+5$

$\Rightarrow 4\le \sin 2x+5\le 6$

$\Rightarrow 4\le h\left(x\right)\le 6$

Hence, Maximum value of $h\left(x\right)$ is $6$ & Minimum value of $h\left(x\right)$ is $4$

4) $f\left(x\right)=|\sin 4x+3|$

We know,

$\sin \theta ϵ[-1,1]$

$\Rightarrow \sin 4xϵ[-1,1]$

$\Rightarrow -1\le \sin 4x\le 1$

$\Rightarrow -1+3\le \sin 4x+3\le 1+3$

$\Rightarrow 2\le \sin 4x+3\le 4$

$\Rightarrow 2\le |\sin 4x+3|\le 4$

$\Rightarrow 2\le f\left(x\right)\le 4$

Hence, Maximum value of $f\left(x\right)$ is $4$ & Minimum value of $f\left(x\right)$ is $2$

5) $h\left(x\right)=x+1,xϵ(-1,1)$
As, $f^{\prime }\left(x\right)=1>0$ (increasing function)

The minimum value occurs, when $x=-1$

The maximum value occurs, when $x=1$

But it’s not possible to locate such points because, $xϵ(-1,1)$

Thus, the given function has neither the maximum value nor minimum value.