Need help finding critical points

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Guest

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I've stumbled across two functions that I would like to find the critical points for, however I'm stuck and honestly don't know how to do it. Any and all help is appreciated. I want to learn how to do it though, so if anyone can post more than just the answer that would be nice.

The first is:
y = (x^2)(x^3 + 1)

the second is:
y = (x - 1)^2(e^x)

While this isn't very urgent, a response within the next day would be great. Thanks in advance!
 
Critical points occur where the derivative (with respect to \(\displaystyle \mbox{x}\)), \(\displaystyle \frac{dy}{dx}\), is equal to zero or is undefined.

\(\displaystyle \mbox{ y = (x^2)(x^3 + 1)}\)

Expand the parentheses:

\(\displaystyle \mbox{ y = x^5 + x^2}\)

Now differentiate term by term:

\(\displaystyle \mbox{ \frac{dy}{dx} = 5x^4 + 2x}\)

Now set \(\displaystyle \mbox{\frac{dy}{dx} = 0}\):

\(\displaystyle \mbox{ 5x^4 + 2x = 0}\)

Factorise to solve for x:

\(\displaystyle \mbox{ x(5x^3 + 2) = 0}\)

Therefore

\(\displaystyle \mbox{ x= 0}\)
\(\displaystyle \mbox{ }\)or \(\displaystyle \mbox{ 5x^3 + 2 = 0 \Rightarrow x^3 = -\frac{2}{5} \Rightarrow x = ^3\sqrt{-\frac{2}{5} } \approx -0.737}\)

The second question should be ok if you're familiar with the product and chain rules (otherwise just a bit tedious).
 
Thank you for the quick response. I will take a stab at the second question as I am somewhat familiar with the chain rule. If I can't get it, though, I will reply again, or edit this one saying that I still need help. Thanks again.
 
Edited posts don't show as new, but replies certainly do.
 
Alright I've tried it, but I can't get it...Can anyone help me with it please?
 
\(\displaystyle \mbox{(x - 1)^2\cdot e^x}\)

We can apply the product rule:

\(\displaystyle \mbox{\frac{d}{dx} u\cdot v = u\cdot \frac{dv}{dx} + v\cdot \frac{du}{dx}}\)

Take
\(\displaystyle \mbox{ u = (x - 1)^2 \Rightarrow \frac{du}{dx} = 2(x - 1)}\)
\(\displaystyle \mbox{ }\)(by the chain rule)

and \(\displaystyle \mbox{ v = e^x \Rightarrow \frac{dv}{dx} = e^x}\)

And we have

\(\displaystyle \mbox{\frac{d}{dx} (x - 1)^2\cdot e^x = (x - 1)^2\cdot e^x + 2(x - 1)\cdot e^x}\)
 
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