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From: Albert on 8 May 2010 08:43
Let f(x) = (x ^ 2) * e ^ x . Find {x: f'(x) < 0}

f'(x) = (x ^ 3) * e ^ x.
Solve (x ^ 3) * e ^ x = 0
x = 0
f'(-1) = -e ^ -1 = -1 / e
f'(1) = e
Sign diagram indicates gradient is negative when x < 0.
Now I graphed f(x) on my calculator; it's clear that the gradient is
not always negative before zero. How do I determine this without a
calculator?

TIA.
From: karl on 8 May 2010 08:52
Albert schrieb:
> Let f(x) = (x ^ 2) * e ^ x . Find {x: f'(x) < 0}
>
> f'(x) = (x ^ 3) * e ^ x.

What is this, the derivative of f(x)? If you think this, read in your textbook about the product rule for
differentiating functions.

> Solve (x ^ 3) * e ^ x = 0
> x = 0
> f'(-1) = -e ^ -1 = -1 / e
> f'(1) = e
> Sign diagram indicates gradient is negative when x < 0.
>
From: Albert on 8 May 2010 08:57
karl wrote:
> Albert schrieb:
> >                 f'(x) = (x ^ 3) * e ^ x.
>
> What is this, the derivative of f(x)? If you think this, read in your textbook about the product rule for
> differentiating functions.

Yes, that is such a poor mistake I made.

f'(x) = xe^x(2 + x)
From: karl on 8 May 2010 09:00
Albert schrieb:
> karl wrote:
>> Albert schrieb:
>>> f'(x) = (x ^ 3) * e ^ x.
>> What is this, the derivative of f(x)? If you think this, read in your textbook about the product rule for
>> differentiating functions.
>
> Yes, that is such a poor mistake I made.
>
> f'(x) = xe^x(2 + x)

I said, read in your textbook about the product rule.
From: amzoti on 8 May 2010 11:05
On May 8, 5:57 am, Albert <albert.xtheunkno...(a)gmail.com> wrote:
> karl wrote:
> > Albert schrieb:
> > >                 f'(x) = (x ^ 3) * e ^ x.
>
> > What is this, the derivative of f(x)? If you think this, read in your textbook about the product rule for
> > differentiating functions.
>
> Yes, that is such a poor mistake I made.
>
> f'(x) = xe^x(2 + x)

If I asked you to find the derivative of u*v, how would you respond?

You can see it here: http://www.math.hmc.edu/calculus/tutorials/prodrule/

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