Question

attempt 1: 10 attempts remaining. find the exact location of all the relative and absolute extrema of the function $f(x)=3x^{2}-12x - 15$ with domain $0,5$. f has? at $(x,y)=($ , $)$ (smallest $x$-value) f has? at $(x,y)=($ , $)$ (largest $x$-value) submit answer next item

Explanation:

Step1: Find the derivative

Differentiate $f(x)=3x^{2}-12x - 15$ using the power - rule. $f^\prime(x)=6x - 12$.

Step2: Find critical points

Set $f^\prime(x)=0$. So, $6x - 12 = 0$. Solving for $x$ gives $x = 2$.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x = 0$, $x = 2$, and $x = 5$.
When $x = 0$, $f(0)=3(0)^{2}-12(0)-15=-15$.
When $x = 2$, $f(2)=3(2)^{2}-12(2)-15=3\times4-24 - 15=12-24 - 15=-27$.
When $x = 5$, $f(5)=3(5)^{2}-12(5)-15=3\times25-60 - 15=75-60 - 15=0$.

Step4: Determine the extrema

Since the second - derivative $f^{\prime\prime}(x)=6>0$, the function is concave up.
The relative and absolute minimum is at $(x,y)=(2,-27)$.
The absolute maximum on the domain $[0,5]$ is at $(x,y)=(5,0)$.

Answer:

$f$ has a relative and absolute minimum at $(x,y)=(2,-27)$ (smallest $x$-value)
$f$ has an absolute maximum at $(x,y)=(5,0)$ (largest $x$-value)