what is the value of $x$ at which the minimum value of $y = 3x^{\frac{4}{3}}-2x$ occurs on the closed…

Explanation:

Step1: Find the derivative of y

Using the power - rule $(x^n)'=nx^{n - 1}$, if $y = 3x^{\frac{4}{3}}-2x$, then $y'=3\times\frac{4}{3}x^{\frac{4}{3}-1}-2=4x^{\frac{1}{3}}-2$.

Step2: Set the derivative equal to 0

Set $y' = 0$, so $4x^{\frac{1}{3}}-2 = 0$. Then $4x^{\frac{1}{3}}=2$, and $x^{\frac{1}{3}}=\frac{1}{2}$. Cubing both sides gives $x=\frac{1}{8}$.

Step3: Evaluate y at critical points and endpoints

• When $x = 0$, $y=3\times0^{\frac{4}{3}}-2\times0 = 0$.
• When $x=\frac{1}{8}$, $y = 3\times(\frac{1}{8})^{\frac{4}{3}}-2\times\frac{1}{8}=3\times(\frac{1}{2})^4-\frac{1}{4}=3\times\frac{1}{16}-\frac{1}{4}=\frac{3 - 4}{16}=-\frac{1}{16}$.
• When $x = 1$, $y=3\times1^{\frac{4}{3}}-2\times1=3 - 2 = 1$.

Answer:

B. $\frac{1}{8}$