Could anyone give me some help in the right direction for solving the following:
Show the equation $\tan\left[\frac{x^2+5x+19}{10}\right]=\ln\left[\frac{5}{x+9}\right]$ has a unique solution in $[0,2]$.
Im assuming I check for a change of sign to see that there is a solution but not sure exactly how to go about this. Also not sure how to show that it is unique.
Thanks in advance.
Consider the function as @gimusi proposed and differentiate it
$$f(x)=\tan\left[\frac{x^2+5x+19}{10}\right]-\ln\left[\frac{5}{x+9}\right]$$ $$=\tan\left[\frac{x^2+5x+19}{10}\right]-\ln 5+\ln \left[x+9 \right] $$
use the intermediate value theorem to show the existence of a solution and show that the derivative
$$f'(x)=\left[1+\tan^2\left[\frac{x^2+5x+19}{10}\right]\right]\dfrac{2x+5}{10}+\dfrac{1}{x+9}$$
has a strict sign (strictly positive or negative) on $x\in I=\left[0,2\right]$. For the derivative of the $\tan$ function I used $\left[ \tan x\right]'=1+\tan^2 x$ and the chain rule.
Note, that the first bracket is greater or equal to $1$. The fraction $\dfrac{2x+5}{10}$ is positive on the interval $I$ and so is the fraction $\dfrac{1}{x+9}$. Hence, the derivative is strictly positive on the interval $I$.