What are the critical values, if any, of #f(x) = f(x) = x^{2}e^{15 x}#?

1 Answer
Nov 4, 2015

Answer:

#xe^{15x} ( 2+15x)#

Explanation:

To find the critical points, we need the first derivative. This function is a multiplication of a power and a composite exponential. Let's see how to deal with these three things:

  • The derivative of a multiplication #f*g# is (Leibniz formula) #f'*g + f*g'#;
  • The derivative of a power #x^n# is #nx^{n-1}#;
  • The derivative of a composite function #f(g(x))# is #f'(g(x)) * g'(x)#;
  • The derivative of an exponential #e^x# is the exponential itself.

Let's put all these things together:

  • The derivative of #x^2 e^{15x}# is

#(d/dx x^2) * e^{15x} + x^2 * (d/dx e^{15x})#

  • The derivative of #x^2 is #2x^1=2x#, and the expression becomes

#2x * e^{15x} + x^2 * (d/dx e^{15x})#

  • The exponential is a composite function, so we must derive the exponential and then multiply for the derivative of the exponent:

#d/dx e^{15x} = e^{15x} * (d/dx 15x) = e^{15x} * 15#

So, the answer is

#2x * e^{15x} + 15x^2 e^{15x}#

We can factor an exponential and a #x#, obtaining

#xe^{15x} ( 2+15x)#