Question

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

Answer 1

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.