Find Minimum and Maximum Values of Functions: Problems with Solutions

Solution:
$f'(x)=4x^3-4x=0$

$4x(x^2-1)=0$
$4x(x-1)(x+1)=0$
$x=0, -1, 1$

	$-3$						$3$
		$-1$		$0$		$1$
$f'(x)$	$-$	$0$	$+$	$0$	$-$	$0$	$+$
$f(x)$	$\searrow$		$\nearrow$		$\searrow$		$\nearrow$

$f_{\text{min}}(x)=f(-1)=(-1)^4-2(-1)^2+5=4$
$f_{\text{min}}(x)=f(1)=(1)^4-2(1)^2+5=4$
$\Rightarrow f_{\text{min}}(x)=f(-1)=f(1)=4$


$f_{\text{max}}(x)=f(-3)=(-3)^4-2(-3)^2+5=68$
$f_{\text{max}}(x)=f(3)=(3)^4-2(3)^2+5=68$
$f_{\text{max}}(x)=f(0)=(0)^4-2(0)^2+5=5$
$\Rightarrow f_{\text{max}}(x)=f(-3)=f(3)=68$


Answer
Minimum value: $4$
Maximum value: $68$

Solution:
$f'(x)=4x^3-4x=0$

$4x(x^2-1)=0$
$4x(x-1)(x+1)=0$
$x=0, -1, 1$

		$-1$		$0$		$1$
$f'(x)$	$-$	$0$	$+$	$0$	$-$	$0$	$+$
$f(x)$	$\searrow$		$\nearrow$		$\searrow$		$\nearrow$

$f_{\text{min}}(x)=f(-1)=(-1)^4-2(-1)^2+5=4$
$f_{\text{min}}(x)=f(1)=(1)^4-2(1)^2+5=4$
$\Rightarrow f_{\text{min}}(x)=f(-1)=f(1)=4$


$f_{\text{max}}(x)=f(0)=(0)^4-2(0)^2+5=5$

Answer
Minimum value: $4$
Maximum value: $5$

Solution:
$f'(x)=[(x-2)^2]'=3(x-2)^2(x-2)'=3(x-2)^2>0$

The function is increasing.

$f_{\text{min}}(x)=f(0)=(0-2)^3=-8$
$f_{\text{max}}(x)=f(2)=(2-2)^3=0$
Answer
Minimum value: $-8$
Maximum value: $0$

Solution:
$f'(x)=3(1-x)^2(1-x)'=3(1-x)^2(-1)=-3(1-x)^2<0$

The function is decreasing.

$f_{\text{min}}=f(1)=(1-1)^3=0$

$f_{\text{max}}=f(-1)=(1-(-1))^3=2^3=8$

Solution:
$f(x)=-x^2-x+2$

$f'(x)=(-x^2-x+2)'=-2x-1$


$f'(x)=0$
$-2x-1 = 0$
$-2x = 1$
$x=-\frac{1}{2}$

	$-1$		$2$
		$-\frac{1}{2}$
$f'(x)$	$+$	$0$	$-$
$f(x)$	$\nearrow$		$\searrow$

$f_{\text{max}}(x)=f(-\frac{1}{2})=-(-\frac{1}{2})^2-(-\frac{1}{2})+2=-\frac{1}{4}+\frac{1}{2}+2=\frac{-1+2+8}{4}=\frac{9}{4}$

$f_{\text{min}}(x)=f(-1)=-(-1)^2-(-1)+2=-1+1+2=2$
$f_{\text{min}}(x)=f(2)=-(2)^2-(2)+2=-4-2+2=-4$
$\Rightarrow f_{\text{min}} (x)=f(2)=-4$

Answer:
Max: $\frac{9}{4}$, Min: $-4$

Solution:
$f'(x)=(x^4-8x^2-9)'=4x^3-2\cdot8x=4x^3-16x$
$f'(x)=0$
$4x^3-16x = 0$
$x^3-4x = 0$
$x(x^2-4)= 0$
$x(x-2)(x+2)= 0$
Critical points: $-2, 0, 2$

		$-2$		$0$		$2$
$f'(x)$	$-$	$0$	$+$	$0$	$-$	$0$	$+$
$f(x)$	$\searrow$		$\nearrow$		$\searrow$		$\nearrow$

Max: $f(0)=0^4-8\cdot0^2-9=-9$
$f(3)=3^4-8\cdot3^2-9=81-72-9=0$
$\Rightarrow $ max: $f(3)=0$
Min: $f(2)=2^4-8\cdot2^2-9=-25$

Answer:
Min: $-25$, Max: $0$

Solution:

	$(-\infin, 3)$	$[3, +\infin)$
$x-3$	-	+

Case 1 $x\in(-\infin,3)$
$f(x)=2(-(x-3))+x^2=2(3-x)+x^2=x^2-2x+6$
$x\in[5, 7]$ but $[5, 7]\notin(-\infin,3)$ So we are not interested in this case.

Case 2 $x\in[3, +\infin)$
$f(x)=2(x-3)+x^2=x^2+2x-6$
$f'(x)=(x^2+2x-6)'=2x+2$
$f'(x)=0$
$2x+2=0$
$x=-1$

In the interval $[3, +\infin)$ the function is increasing.
Min: $f(5)=5^2+2\cdot5-6=29$
Max: $f(7)=7^2+2\cdot7-6=57$

Answer: $29$