Let $f ( x ) = 2 x ^ { 3 } - 15 x ^ { 2 } + 36 x + 5$ The set of all critical numbers of ƒ is

A) $\{ - 2 , - 3 \}$
B) $\{ - 2,3 \}$
C) $\{ 2 , - 3 \}$
D) $\{ 2,3 \}$
E) $\{ 1,3 \}$

Question

Let $f ( x ) = 2 x ^ { 3 } - 15 x ^ { 2 } + 36 x + 5$ The set of all critical numbers of ƒ is

A) $\{ - 2 , - 3 \}$
B) $\{ - 2,3 \}$
C) $\{ 2 , - 3 \}$
D) $\{ 2,3 \}$
E) $\{ 1,3 \}$

Quizplus · Accepted Answer

To find the critical numbers of $f(x)$, we first find its derivative $f'(x)$ and then solve $f'(x) = 0$. The derivative is $f'(x) = 6x^2 - 30x + 36$. Factoring or using the quadratic formula, we find the roots of $f'(x) = 0$ to be $x = 2$ and $x = 3$, which are the critical numbers of $f$.

Let f(x)=2x3−15x2+36x+5f ( x ) = 2 x ^ { 3 } - 15 x ^ { 2 } + 36 x + 5f(x)=2x3−15x2+36x+5

Let $f ( x ) = 2 x ^ { 3 } - 15 x ^ { 2 } + 36 x + 5$