Find the critical numbers of $f ( x ) = x ^ { 4 } ( x - 3 ) ^ { 3 }$.
A) $0,3 , \frac { 12 } { 7 }$
B) $0,4 , \frac { 7 } { 12 }$
C) $0,4 , \frac { 12 } { 7 }$
D) $0,3 , \frac { 12 } { 11 }$
E) $0,3 , \frac { 7 } { 12 }$

Question

Quizplus · Accepted Answer

To find the critical numbers, we need to find the derivative $f'(x)$ and set it to zero. The derivative is $f'(x) = 4x^3(x-3)^3 + 3x^4(x-3)^2(-1)$. Setting $f'(x) = 0$ gives critical points at $x = 0$, $x = 3$, and solving the equation $4(x-3) = -3x$ gives $x = \frac{12}{7}$.

Find the Critical Numbers Of f(x)=x4(x−3)3f ( x ) = x ^ { 4 } ( x - 3 ) ^ { 3 }f(x)=x4(x−3)3

Find the Critical Numbers Of $f ( x ) = x ^ { 4 } ( x - 3 ) ^ { 3 }$