consider the function f(x)=81x^(1/3)-3x. over the domain 0,140. a) list and classify any critical values…

Explanation:

Step1: Find the derivative of the function

The derivative of $y = 81x^{\frac{1}{3}}-3x$ using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ is $y'=81\times\frac{1}{3}x^{-\frac{2}{3}}-3=27x^{-\frac{2}{3}}-3=\frac{27}{x^{\frac{2}{3}}}-3$.

Step2: Set the derivative equal to zero to find critical points

Set $\frac{27}{x^{\frac{2}{3}}}-3 = 0$. Add 3 to both sides: $\frac{27}{x^{\frac{2}{3}}}=3$. Cross - multiply to get $27 = 3x^{\frac{2}{3}}$. Then divide both sides by 3: $x^{\frac{2}{3}} = 9$. Raise both sides to the $\frac{3}{2}$ power. We have $x=\pm27$. But our domain is $[0,140]$, so we only consider $x = 27$. Also, the derivative is undefined when $x = 0$ (since the denominator of $y'=\frac{27}{x^{\frac{2}{3}}}-3$ is zero at $x = 0$). So the critical values in the domain $[0,140]$ are $x = 0$ and $x=27$. For $x = 0$, the derivative $y'$ is undefined. For $x = 27$, $y'=0$.

Step3: Classify the critical points

We can use the first - derivative test. Consider the intervals $[0,27)$ and $(27,140]$. Take a test point in the interval $[0,27)$, say $x = 1$. Then $y'(1)=\frac{27}{1^{\frac{2}{3}}}-3=27 - 3=24>0$. Take a test point in the interval $(27,140]$, say $x = 64$. Then $y'(64)=\frac{27}{64^{\frac{2}{3}}}-3=\frac{27}{16}-3=\frac{27 - 48}{16}<0$. Since the function is increasing on $[0,27)$ and decreasing on $(27,140]$, at $x = 27$ the function has a local maximum. At $x = 0$, the function has a non - differentiable point.

Answer:

a) Critical values: $x = 0,x = 27$. Classification: $x = 0$ is a non - differentiable point, $x = 27$ is a local maximum. b) At $x = 0$, it is a non - differentiable point.