Question

determine all critical points for the function.
y = 4x² - 128√x
a. x = 0, x = 4, and x = -4
b. x = 0
c. x = 0 and x = 4
d. x = 4

Explanation:

Step1: Find the derivative of the function

The function is \(y = 4x^{2}-128\sqrt{x}=4x^{2}-128x^{\frac{1}{2}}\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(y^\prime=8x-\frac{128}{2}x^{-\frac{1}{2}}=8x-\frac{64}{\sqrt{x}}\).

Step2: Set the derivative equal to zero

Set \(y^\prime = 0\), so \(8x-\frac{64}{\sqrt{x}}=0\).
Multiply through by \(\sqrt{x}\) (since \(x\geq0\) for the original function \(y = 4x^{2}-128\sqrt{x}\)) to get \(8x^{\frac{3}{2}}-64 = 0\).
Then \(8x^{\frac{3}{2}}=64\), \(x^{\frac{3}{2}} = 8\), and \(x=8^{\frac{2}{3}}=(2^{3})^{\frac{2}{3}}=4\).
Also, consider where the derivative is undefined. The derivative \(y^\prime=8x-\frac{64}{\sqrt{x}}\) is undefined at \(x = 0\) (since \(\frac{1}{\sqrt{x}}\) is undefined at \(x = 0\)).

Answer:

C. \(x = 0\) and \(x = 4\)