# How do you find all critical numbers of the function f(x)=x^(4/5)(x-4)^2?

Dec 27, 2017

$x = 4 , x = \frac{8}{7} , x = 0$

#### Explanation:

Critical points are points in the domain where the derivative is equal to zero or where the derivative is not defined.

Derivative Equal to Zero:

The derivative can be found by using the power rule and the chain rule.

$f ' \left(x\right) = \setminus \frac{14 {x}^{2} - 72 x + 64}{5 {x}^{\setminus \frac{1}{5}}}$

Now we set this equal to zero:

$\setminus \frac{14 {x}^{2} - 72 x + 64}{5 {x}^{\setminus \frac{1}{5}}} = 0$

$14 {x}^{2} - 72 x + 64 = 0$

$2 \left(7 {x}^{2} - 36 x + 32\right) = 0$

$2 \left(7 x - 8\right) \left(x - 4\right) = 0$

We get:

$7 x - 8 = 0 \rightarrow x = \frac{8}{7}$
$x - 4 = 0 \rightarrow x = 4$

Derivative Not Defined:

The only possible case where $\setminus \frac{14 {x}^{2} - 72 x + 64}{5 {x}^{\setminus \frac{1}{5}}}$ is not defined is when the denominator is equal to zero.

Therefore we set $5 {x}^{\frac{1}{5}} = 0$ and solve.

$5 {x}^{\frac{1}{5}} = 0$

$x = 0$

Therefore, the derivative is not defined at $x = 0$.

$0$ is in the domain of $f$ so $0$ is a critical point for $f$.

Conclusion

The critical points of this function are $x = 4 , x = \frac{8}{7} , x = 0$