Identify the function's local and absolute extreme values, if any, saying where they occur.
-$$f ( x ) = - x ^ { 3 } + 1.5 x ^ { 2 } + 36 x + 2$$
A) local maximum at x = -4; local minimum at x = 3
B) local maximum at x = -3; local minimum at x = 4
C) local maximum at x = 4; local minimum at x = -3
D) local maximum at x = 3; local minimum at x = -4

Question

Quizplus · Accepted Answer

To find the local and absolute extreme values, we take the derivative of the function: 
$$f'(x) = -3x^2 + 3x + 36$$

Setting the derivative equal to zero, we get $-3x^2+3x+36=0$, which simplifies to $x^2-x-12=0$. Factoring this, we get $(x-4)(x+3)=0$, so the derivative equals zero at $x=4$ and $x=-3$.

To determine the nature of the critical points, we can use the second derivative test. 
$$f''(x) = -6x + 3$$

For $x=-3$, $f''(-3) = 15 >0$, so the critical point at $x=-3$ is a local minimum. 
For $x=4$, $f''(4) = -15 <0$, so the critical point at $x=4$ is a local maximum.

To find the absolute maximum and minimum, we can evaluate the function at the critical points as well as the endpoints of the interval $[-5,5]$ (since the degree of the polynomial is odd, the function will approach negative infinity as $x$ approaches negative infinity and positive infinity as $x$ approaches positive infinity).

$f(-5) = 72$, $f(5) = -8$
$f(-3) = 47.25$, $f(4) = 38$
$f(-4) = 82$, $f(3) = 10.5$

Therefore, the absolute maximum value is 82, which occurs at $x=-4$, and the absolute minimum value is -8, which occurs at $x=5$.

The only choice that matches these results is C.

Identify the Function's Local and Absolute Extreme Values, If Any f(x)=−x3+1.5x2+36x+2f ( x ) = - x ^ { 3 } + 1.5 x ^ { 2 } + 36 x + 2f(x)=−x3+1.5x2+36x+2

Identify the Function's Local and Absolute Extreme Values, If Any $f ( x ) = - x ^ { 3 } + 1.5 x ^ { 2 } + 36 x + 2$