Question

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)

f(x)=x^{4 / 5}(x - 3)^{2}

x=

resources

Explanation:

Step1: Find the derivative using the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = x^{4/5}\) and \(v=(x - 3)^2\).

First, find \(u^\prime\): Using the power rule \((x^n)^\prime=nx^{n - 1}\), \(u^\prime=\frac{4}{5}x^{-1/5}\).

Second, find \(v^\prime\): Using the chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\), where \(f(u)=u^2\) and \(u = x - 3\). So \(v^\prime = 2(x - 3)\cdot1=2(x - 3)\).

Then \(F^\prime(x)=\frac{4}{5}x^{-1/5}(x - 3)^2+x^{4/5}\cdot2(x - 3)\).

Step2: Simplify the derivative

Factor out \(\frac{2}{5}x^{-1/5}(x - 3)\):

$$
LATEXBLOCK0
$$

Step3: Set the derivative equal to zero

\(F^\prime(x)=0\) when:

• \(x^{-1/5}=0\) (no solution since \(x^{-1/5}=\frac{1}{x^{1/5}}\), and \(\frac{1}{x^{1/5}} = 0\) has no solution for \(x\in R\))

• \(x - 3=0\), then \(x = 3\)

• \(7x-6=0\), then \(x=\frac{6}{7}\)

Also, \(F^\prime(x)\) is undefined when \(x = 0\) (because of the \(x^{-1/5}\) term, \(x^{1/5}=0\) when \(x = 0\))

Answer:

\(0,\frac{6}{7},3\)