Locate the absolute extrema of the function $f ( x ) = 3 x ^ { 2 } + 12 x + 3 \text { on the closed }$ $\text { interval } [ - 4,4 ]$
A) absolute max: $( 4,99 )$; no absolute min
B) no absolute max or min
C) no absolute max; absolute min: $( 4,99 )$
D) absolute max: $( 4,99 )$; absolute min: $( - 2 , - 9 )$
E) absolute max: $( - 2 , - 9 )$; absolute min: $( 4,99 )$

Question

Locate the absolute extrema of the function $f ( x ) = 3 x ^ { 2 } + 12 x + 3 \text { on the closed }$ $\text { interval } [ - 4,4 ]$ 
A) absolute max: $( 4,99 )$; no absolute min
B) no absolute max or min
C) no absolute max; absolute min: $( 4,99 )$
D) absolute max: $( 4,99 )$; absolute min: $( - 2 , - 9 )$
E) absolute max: $( - 2 , - 9 )$; absolute min: $( 4,99 )$

Quizplus · Accepted Answer

To find the absolute extrema of the function, we need to check the critical points and the endpoints of the interval.

First, we find the critical points by finding the derivative of the function and solving for when it equals zero:

$$f'(x) = 6x + 12$$
$$6x + 12 = 0$$
$$x = -2$$

So the only critical point is at $-2$.

Next, we evaluate the function at the critical point and the endpoints of the interval:

$$f(-4) = 51$$
$$f(-2) = -9$$
$$f(4) = 99$$

So the absolute minimum occurs at $(-2,-9)$ and the absolute maximum occurs at $(4,99)$.

Therefore, the answer is choice D.

Locate the Absolute Extrema of the Function f(x)=3x2+12x+3 on the closed f ( x ) = 3 x ^ { 2 } + 12 x + 3 \text { on the closed }f(x)=3x2+12x+3 on the closed

Locate the Absolute Extrema of the Function $f ( x ) = 3 x ^ { 2 } + 12 x + 3 \text { on the closed }$