Expert: Scotto - 5/11/2009

Question
Sorry I should have added this to my previous question because they are similar..

my question is how to I get the derivative of this, so that I can find the critical numbers??
f(x)=(x^2)(e^14x)


Also, how do I know when f(x) is decreasing/increasing??

What I would do for the derivative is first use product law, then what do I use for the e^14x because it is a product law again (I think). Could you run me through it?

I also need help setting it up so that I can find the critical numbers, usually once I find the derivative it's hard for me to clean up the f'(x).

Answer
Let g(x) = x².  Then g'(x) = 2x.
Let h(x) = e^(14x).  Then h'(x) = (d(14x)/dx)e^(14x) = 14e^(14x).
We then can say the f'(x) = g'(x)h(x) + h'(x)g(x).

It is increasing when the derivatve is positive and decreasing when the derivative is negative.

We know that f'(x) = (2x)(e^(14x)) + (14e^(14x))x².
Clearly, we can factor e^(14x) out of both terms.
The result is e^(14x)(2x + 14x²).

Once this has been done,
it can be seen that we can also factor out 2x.
2xe^(14x)(1 + 7x).

So the critical points are where 14x = 0 abd 1 + 7x = 0.
Note that 2e^(14x) is never 0.

The first critical point is x = 0 and the second critical point is where 7x = -1 => x = -1/7.

The intervals to test are then (-∞, -1/7), (-1/7, 0), and (0, ∞).
I would choose -1 in the first interval and 1 in the last interval to test in f'(x), to see what the sign was.  In (-1/7, 0), since it is (-0.142857..., 0), the point to pick might be -0.1.