Question

find the exact location of all the relative and absolute extrema of the function $f(x) = 4x^2 - 16x - 20$ with domain $0, 5$.
$f$ has ? at $(x,y) = (\boxed{\phantom{\
ule{0.5em}{0ex}}},\boxed{\phantom{\
ule{0.5em}{0ex}}})$ (smallest $x$-value)
$f$ has ? at $(x,y) = (\boxed{\phantom{\
ule{0.5em}{0ex}}},\boxed{\phantom{\
ule{0.5em}{0ex}}})$ (largest $x$-value)
next item

Explanation:

Step1: Find critical point

First, take the derivative of $f(x)$:
$f'(x) = \frac{d}{dx}(4x^2 - 16x - 20) = 8x - 16$
Set $f'(x)=0$:
$8x - 16 = 0$
Solve for $x$:
$8x = 16 \implies x=2$

Step2: Evaluate f at critical point

Calculate $f(2)$:
$f(2) = 4(2)^2 - 16(2) - 20 = 4(4) - 32 - 20 = 16 - 32 - 20 = -36$

Step3: Evaluate f at domain endpoints

Calculate $f(0)$:
$f(0) = 4(0)^2 - 16(0) - 20 = -20$
Calculate $f(5)$:
$f(5) = 4(5)^2 - 16(5) - 20 = 4(25) - 80 - 20 = 100 - 80 - 20 = 0$

Step4: Classify extrema

Compare values: $f(2)=-36$ (minimum), $f(5)=0$ (maximum), $f(0)=-20$.
The smallest $x$-value with an extremum is $x=0$ (relative endpoint value), $x=2$ is the absolute minimum. The largest $x$-value with an extremum is $x=5$ (absolute maximum).

Answer:

$f$ has an absolute minimum at $(x,y) = (2, -36)$ (smallest $x$-value for the global minimum; the left endpoint $(0, -20)$ is a relative maximum)
$f$ has an absolute maximum at $(x,y) = (5, 0)$ (largest $x$-value)

To match the structure:
For smallest $x$-value: $f$ has a relative maximum at $(x,y) = (0, -20)$, and an absolute minimum at $(x,y)=(2,-36)$
For largest $x$-value: $f$ has an absolute maximum at $(x,y) = (5, 0)$

Final precise fill-in:

• (smallest $x$-value): $f$ has an absolute minimum at $(x,y) = \boldsymbol{(2, -36)}$; the left endpoint $(0,-20)$ is a relative maximum
• (largest $x$-value): $f$ has an absolute maximum at $(x,y) = \boldsymbol{(5, 0)}$

Primary extrema:
Absolute minimum: $(2, -36)$
Absolute maximum: $(5, 0)$
Relative maximum at left endpoint: $(0, -20)$