Find the critical numbers of $f(x)=x^{4}(x-2)^{3}$ .
A) 0, 3, $\frac{7}{8}$
B) 0, 2, $\frac{12}{11}$
C) $0,2, \frac{8}{7}$
D) 0, 3, $\frac{12}{7}$
E) 0, 2, $\frac{7}{8}$

Question

Quizplus · Accepted Answer

The critical numbers are found by setting the derivative $f'(x)$ equal to zero and solving for $x$. The derivative is $f'(x) = 4x^3(x-2)^3 + 3x^4(x-2)^2(-1)$. Setting this equal to zero and solving gives the critical numbers $x = 0, 2, \frac{8}{7}$.

Find the Critical Numbers Of f(x)=x4(x−2)3f(x)=x^{4}(x-2)^{3}f(x)=x4(x−2)3 A) 0, 3 78\frac{7}{8}87​ B) 0, 2

Find the Critical Numbers Of $f(x)=x^{4}(x-2)^{3}$
A) 0, 3 $\frac{7}{8}$ B) 0, 2