I'm trying to sketch the function $$f(x)=\frac{\ln(5x^2+x)}{2-3x}$$
For the first derivative I get $$\frac{d}{dx}\bigg(\frac{\ln(5 x^2 + x)}{2 - 3 x}\bigg) = \frac{10 x + 1}{(2 - 3 x) (5 x^2 + x)}+\frac{3\ln(5 x^2 + x)}{(2 - 3 x)^2}$$
Then I wanted to solve the inequality $ f'(x)>0 $ (where $f'(x)$ is the first derivative) to see if there is any max/min point, and then I would like to solve $ f''(x)>0 $ to see if the concavity is convex or concave.
The problem is I couldn't solve both $ f'(x)>0 $ and $ f''(x)>0 $ .
It's 3 days I'm looking for a way to solve it and I understood that it is a Transcendental Function and that there are some way to approximate a solution with e.g. Newton's Method (which is pretty cool), but it is a bit convoluted since i should also find the third derivative in order to solve the second derivative inequality.
My question is: Is there an easy way to find max, min and concavity of a function where its first and second derivative are Transcendental functions?
Thanks.
You won't be able to find a closed form expression for the point in $(0, 2/3)$ where the concavity changes, but you probably don't need to. It should be enough to note that $f(x) \to -\infty$ as $x \to 0+$ and $f(x) \to +\infty$ as $x \to 2/3-$, and $f'(x) > 0$ for $0 < x < 2/3$ (note that both numerators and both denominators are positive).