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I'm trying to sketch the function $$f(x)=\frac{\ln(5x^2+x)}{2-3x}$$ .
When I do 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 for any hint

Stramike
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  • I didn't understand what is the issue with Newton Iterations here? Seems like the right way to go. At any rate, this question might be more suited to math.stackexchange – Amir Sagiv Nov 17 '19 at 17:34
  • Newton's method requires to have the derivative of the function you want to approximate. Since I have to find a solution for both first and second derivative, I will need to have second and third derivative to compute it, but computing the third derivative is very time consuming, not mentioning the iterations with Newton's method will be very long. It seems to me that is not a very elegant solution. I wanted to see if there is any other elegant/easy method to solve that transcendental inequality , or to get the min-max points and concavity of the initial function. – Stramike Nov 17 '19 at 17:46
  • hint: put the derivative on a same (positive) denominator, and study the numerator (it is a simpler function) to see when it becomes zero, positive, negative. As said, math.stackexchange is appropriated for such questions. – ama Nov 17 '19 at 20:45

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