# What are the critical values, if any, of f(x)=x^(4/5) (x − 3)^2?

Mar 11, 2016

They are $0$, $3$, and $\frac{6}{7}$.

#### Explanation:

f(x)=x^(4/5) (x − 3)^2

Critical value of $f$ are values in the domain of $f$ at which either $f '$ does not exist or $f ' \left(x\right) = 0$

The domain of $f$ is $\mathbb{R}$.

So every point at which $f '$ fails to exist or $f ' \left(x\right) = 0$ is a critical value for $f$.

Differentiate using the product rule:

$f ' \left(x\right) = \frac{4}{5} {x}^{- \frac{1}{5}} {\left(x - 3\right)}^{2} + {x}^{\frac{4}{5}} 2 \left(x - 3\right) \left(1\right)$ $\text{ }$ (using the chain rule at the end)

$f ' \left(x\right) = \frac{4 {\left(x - 3\right)}^{2}}{5 \sqrt{x}} + \frac{2 {\sqrt{x}}^{4} \left(x - 3\right)}{1}$

$= \frac{4 {\left(x - 3\right)}^{2} + 10 x \left(x - 3\right)}{5 \sqrt{x}}$

$= \frac{2 \left(x - 3\right) \left[2 \left(x - 3\right) + 5 x\right]}{5 \sqrt{x}}$

$= \frac{2 \left(x - 3\right) \left(7 x - 6\right)}{5 \sqrt{x}}$

$f ' \left(0\right)$ does not exist, but $0$ is in the domain of $f$, so $0$ is a critical value for $f$.

$f ' \left(x\right) = 0$ at $x = 3$ and at $x = \frac{6}{7}$, both of which are in the domain of $f$.

The critical values are $0$, $3$, and $\frac{6}{7}$.

We have finished the problem without looking at the graph of $f$. Having done that, it can be helpful to look at the graph.
You can zoom in and out and drag the graph around using a mouse. (It will start the same every time you return to this answer.)

graph{y=x^(4/5)(x-3)^2 [-3.01, 6.857, -0.442, 4.49]}