Question

Find the maximum and minimum values, if any, of the following functions given by

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

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

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

$f\left(x\right)=|sin4x+3|$

$h\left(x\right)=x+1,xϵ(-1,1)$

Answer 1

Given, functions is $f\left(x\right)=|x+2|-1$
We know that $|x+2|\ge 0forallxϵR$
Therefore, $f\left(x\right)=|x+2|-1\ge -1foreveryxϵR$
The minimum value of f is attained when $|x+2|=0$
i.e., $|x+2|=0\Rightarrow x=-2$
$\therefore$ Minimum value of $f=f(-2)=|-2+2|-1=0-1=-1$
Hence, f(x) has minimum value -1 at x=2 but f(x) has no maximum value.

Note: The modulus of a function is always greater than or equal to 0.

Given function is, $g\left(x\right)=-|x+1|+3$
We know that $|x+1|\ge 0$ for all $xϵR$
$\Rightarrow -|x-1|\le 0$ for all $xϵR\Rightarrow -|x+1|+3\le 3$ for all $xϵR$
The maximum value of g is attained when |x+1| =0
i.e., $|x+1|=0\Rightarrow x=-1$
$\therefore$ Maximum value of $g=g(-1)=-|-1+1|+3=3$
Hence, g(x) has maximum value is at x=-1 but g(x) has no minimum value.

Given function is, $h\left(x\right)=sin2x+5$
We know that, $-1\le sinx\le 1\Rightarrow -1\le sin2x\le 1$
$\Rightarrow -1+5\le sin2x+5\le 1+5\Rightarrow 4\le sin2x+5\le 6$
Hence, maximum value of h(x) is 6 and minimum value of h(x) is 4.

Given function is, $f\left(x\right)=|sin4x+3|$
We know that $-1\le sin4x\le 1\Rightarrow 3-1\le sin4x+3\le 1+3$
$\Rightarrow 2\le sin4x+3\le 4\Rightarrow 2\le |sin4x+3|\le 4$
Hence, maximum value of f(x) is 4 and minimum value of f(x) is 2.

Given function is, $h\left(x\right)=x+1,-1<x<1$
Now $-1<x<1\iff -1+1<x+1<1+1\iff 0<x+1<2$
Hence, on the point (0,2), h(x) has neither a maximum nor a minimum value.
Alternate method
h'(x)=1 exists at all $xϵ(-1,1)$
Also, $h^{\prime }\left(x\right)\ne 0$ for any $xϵ(-1,1)$
Further h(x) is defined on an open interval.
So, there is no point for which f(x) may have a minimum or a maximum value.