Let $f ( x ) = x ^ { 4 } - 4 x ^ { 3 } + 16 x$ Then ƒ has a point of inflection at x =

A) $- \frac { 1 } { 3 } , 0$
B) $1,0$
C) $\frac { 1 } { 3 } , 0$
D) $- 1,0$
E) $2,0$

Question

Let $f ( x ) = x ^ { 4 } - 4 x ^ { 3 } + 16 x$ Then ƒ has a point of inflection at x =

A) $- \frac { 1 } { 3 } , 0$
B) $1,0$
C) $\frac { 1 } { 3 } , 0$
D) $- 1,0$
E) $2,0$

Quizplus · Accepted Answer

A point of inflection occurs where the second derivative changes sign. The second derivative of $f(x)$ is $f''(x) = 12x^2 - 24x$. Setting $f''(x) = 0$ gives $x(x-2) = 0$, so $x = 0$ or $x = 2$. Testing intervals around these points shows a sign change at $x = 2$.

Let f(x)=x4−4x3+16xf ( x ) = x ^ { 4 } - 4 x ^ { 3 } + 16 xf(x)=x4−4x3+16x

Let $f ( x ) = x ^ { 4 } - 4 x ^ { 3 } + 16 x$