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

Nov 10, 2015

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 $\ln \left(6 {x}^{2} + 6 x\right)$, 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 $\frac{1}{6 {x}^{2} + 6 x}$. This must be multiplied by the derivative of $6 {x}^{2} + 6 x$, which is $12 x + 6$

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

$f ' \left(x\right) = {e}^{x} + \frac{12 x + 6}{6 {x}^{2} + 6 x} = {e}^{x} + \frac{2 x + 1}{{x}^{2} + x}$

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

${e}^{x} + \frac{2 x + 1}{{x}^{2} + x} = 0$

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.