Determine all critical points for the function.
-$$y = 4 x ^ { 2 } - 128 \sqrt { x }$$
A) $x = 0$
B) $x = 4$
C) $x = 0$ and $x = 4$
D) $x = 0 , x = 4$, and $x = - 4$

Question

Quizplus · Accepted Answer

To find the critical points of the function $y = 4x^2 - 128\sqrt{x}$, we first need to find its derivative and then set that derivative equal to zero. The derivative of $y$ with respect to $x$ is:$$y' = \frac{d}{dx}(4x^2) - \frac{d}{dx}(128\sqrt{x}) = 8x - \frac{128}{2\sqrt{x}} = 8x - \frac{64}{\sqrt{x}}.$$Setting $y'$ equal to zero gives:$$8x - \frac{64}{\sqrt{x}} = 0.$$Multiplying through by $\sqrt{x}$ to clear the denominator gives:$$8x\sqrt{x} - 64 = 0.$$$$8x\sqrt{x} = 64.$$$$x\sqrt{x} = 8.$$$$x^{\frac{3}{2}} = 8.$$$$x = 8^{\frac{2}{3}}.$$$$x = 4.$$So, $x = 4$ is a critical point. Additionally, we must consider the domain of the original function. Since the function includes $\sqrt{x}$, $x$ must be greater than or equal to 0. Thus, $x = 0$ is also a critical point because it's a boundary of the domain where the derivative does not exist. Therefore, the correct answer is that the critical points are $x = 0$ and $x = 4$.

Determine All Critical Points for the Function y=4x2−128xy = 4 x ^ { 2 } - 128 \sqrt { x }y=4x2−128x​

Determine All Critical Points for the Function $y = 4 x ^ { 2 } - 128 \sqrt { x }$