# Thread: Very Simple Critical Points Problem with Answer (Need help understanding)

1. ## Very Simple Critical Points Problem with Answer (Need help understanding)

Find the critical numbers of the function

x^(4/5) * ((x-4)²)

I used the chain rule. Got 8/5x^(-1/5) * (x-4).

That would give me critical points of 0 and 4.

Where does this 8/7 value come from?

Thanks again guys!!

Also I'm sure many of you are about to watch football so good luck!! (Go Skins!!)

2. ## Re: Very Simple Critical Points Problem with Answer (Need help understanding)

Using the product rule, we find:

$\displaystyle f'(x)=x^{\frac{4}{5}}(2(x-4))+\frac{4}{5}x^{-\frac{1}{5}}(x-4)^2=$

$\displaystyle \frac{1}{5}x^{-\frac{1}{5}}(x-4)(10x+4(x-4))=\frac{2}{5}x^{-\frac{1}{5}}(x-4)(7x-8)$

3. ## Re: Very Simple Critical Points Problem with Answer (Need help understanding)

Originally Posted by MarkFL2
Using the product rule, we find:

$\displaystyle f'(x)=x^{\frac{4}{5}}(2(x-4))+\frac{4}{5}x^{-\frac{1}{5}}(x-4)^2=$

$\displaystyle \frac{1}{5}x^{-\frac{1}{5}}(x-4)(10x+4(x-4))=\frac{2}{5}x^{-\frac{1}{5}}(x-4)(7x-8)$

I'm sooo confused. Why on Earth would you use the product rule? Maybe I wrote it wrong. it looks like

X^4/5(x-4)²

That looks like classic chain rule? Why wouldn't you use the chain rule. (Sorry if I wrote it wrong!!)

4. ## Re: Very Simple Critical Points Problem with Answer (Need help understanding)

You can use the product rule because the given function is the product of $\displaystyle x^{\frac{4}{5}}$ and $\displaystyle (x-4)^2$.

5. ## Re: Very Simple Critical Points Problem with Answer (Need help understanding)

To explain a little further, you use the chain rule for compositions - functions of the form f(g(x)), and the product rule for products - functions of the form f(x)g(x). For most functions you don't get to choose; you have to use whichever is appropriate.

The function you have is a product, so you use the product rule:

$\displaystyle f(x)=(x^{\frac{4}{5}})((x-4)^2)$
$\displaystyle f(x)=(x^{\frac{4}{5}})((x-4)^2)'+(x^{\frac{4}{5}})'((x-4)^2)$

You actually use the chain rule for $\displaystyle (x-4)^2$, though you probably do it without really thinking. It is the composition of $\displaystyle f(y)=y^2$ and $\displaystyle g(x)=x-4$. So $\displaystyle f(g(x))=(x-4)^2$ and by the chain rule, $\displaystyle (f(g(x)))'=f'(g(x))g'(x)=(2(x-4))(1)$.

and then what MarkFL2 said.

- Hollywood