Find the absolute maximum of the function $f ( x ) = x ^ { 3 }$ on the interval $- \frac { 1 } { 2 } \leq x \leq 1$ .
A) 0
B) $\frac { 1 } { 8 }$
C) -1
D) 1

Question

Quizplus · Accepted Answer

The function $f(x) = x^3$ is a cubic function, which is continuous and differentiable everywhere. To find the absolute maximum on the interval $-\frac{1}{2} \leq x \leq 1$, we evaluate the function at the endpoints of the interval and at any critical points within the interval. The derivative of $f(x)$ is $f'(x) = 3x^2$, which is always non-negative, meaning the function is always increasing on the interval. Therefore, the absolute maximum occurs at the right endpoint of the interval, which is $f(1) = 1^3 = 1$.

Find the Absolute Maximum of the Function f(x)=x3f ( x ) = x ^ { 3 }f(x)=x3

Find the Absolute Maximum of the Function $f ( x ) = x ^ { 3 }$