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 of $f(x) = x^4(x - 3)^3$, we first find its derivative: $f'(x) = 4x^3(x - 3)^3 + 3x^4(x - 3)^2$. Setting $f'(x) = 0$ to find critical points, we factor to get $x^3(x - 3)^2(4(x - 3) + 3) = 0$, which simplifies to $x^3(x - 3)^2(7x - 12) = 0$. Solving for $x$, we get $x = 0$, $x = 3$, and $x = \frac{12}{7}$, which are the critical numbers.

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 $f ( x ) = x ^ { 4 } ( x - 3 ) ^ { 3 }$