What are the critical points of #f (x) = e^x + ln(6x^2+x)#?

1 Answer
Nov 10, 2015

Answer:

An approximate answer is #x=-0.419971#

Explanation:

To find the critical points, we need to compute the first derivative. Since the derivative of a sum is the sum of the derivatives, we can split the problem in two subproblems:

  • The derivative of #e^x# is simply #e^x# itself, so the first term is easy to solve

  • As for #ln(6x^2+6x)#, we need to use the chain rule: we have to differentiate the outer function, and then multiply for the derivative of the inner function. The outer function is a logarithm, and so its derivative is the inverse of the argument, which is #1/(6x^2+6x)#. This must be multiplied by the derivative of #6x^2+6x#, which is #12x+6#

Now we have to sum the two terms to obtain the derivative:

#f'(x)= e^x + (12x+6)/(6x^2+6x) = e^x + (2x+1)/(x^2+x)#

The critical points are the zeroes of the derivative, so we should solve

#e^x + (2x+1)/(x^2+x)=0#

but this is a trascendental equation, so the best you can do is asking a calculator for an approximate value of the solution, as for example here.