Consider the function f(x) = x^{2}e^{10 x}.?

Question

Find the first derivative of f(x). Enter exp(10 x) when you write the exponent.

f'(x)=

For the function f(x) there are two critical numbers, which correspond to a local maximum and a local minimum.
The x-coordinate of the LMAX is
The x-coordinate of the LMIN is

Answer 1

f'(x)=(10x^2+2x)e^(10x) because of product rule.

set it equal to zero to find the max and min.

0=(10x^2+2x)e^(10x)


0=2x(5x+1) {e^10x went to zero)

2x=0 5x+1=0
x=0 x=-1/5

plug those back in.


to find LMAX, it has to be less than 0 to make it concave down giving a local max.

to find LMIN, it has to be greater than 0 to make it concave up giving a local min.

hope this helped.

Answer 2

f(x) = (x^2)e^(10 x)
f'(x) = 10(x^2)e^(10x) + 2xe^(10x)
f'(x) = 2xe^(10x)(5x + 1) = 0 for local max & min
The critical points are x = - 1/5, 0
The x-coordinate of the LMAX is - 1/5
f(-1/5) = (1/25)e^-2 = 0.005413411
The x-coordinate of the LMIN is 0
f(0) = 0