Find the critical points of $f ( x ) = e ^ { x } \left( x ^ { 2 } - 2 x \right)$
A) $\sqrt { 2 }$
B) - 2, 2
C) - , $\sqrt { 2 }$ $\sqrt { 2 }$
D) 2

Question

Quizplus · Accepted Answer

To find the critical points of $f(x) = e^x(x^2 - 2x)$, we first need to find its derivative and then set that derivative equal to zero. The derivative, using the product rule, is $f'(x) = e^x(x^2 - 2x) + e^x(2x - 2) = e^x(x^2 - 2x + 2x - 2) = e^x(x^2 - 2)$. Setting $f'(x) = 0$ gives $e^x(x^2 - 2) = 0$. Since $e^x$ is never zero, we only need to solve $x^2 - 2 = 0$, which gives $x^2 = 2$ or $x = \pm \sqrt{2}$.

Find the Critical Points Of f(x)=ex(x2−2x)f ( x ) = e ^ { x } \left( x ^ { 2 } - 2 x \right)f(x)=ex(x2−2x)

Find the Critical Points Of $f ( x ) = e ^ { x } \left( x ^ { 2 } - 2 x \right)$