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

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Answer 1 · 2017-12-27T23:41:17

$x=4, x=8/7,x=0$

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.

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

Now we set this equal to zero:

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

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

$2(7x^2-36x+32)=0$

$2(7x-8)(x-4)=0$

We get:

$7x-8=0rarrx=8/7$
$x-4=0rarrx=4$

Derivative Not Defined:

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

Therefore we set $5x^(1/5)=0$ and solve.

$5x^(1/5)=0$

$x=0$

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 $x=4, x=8/7,x=0$

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