Given $y = f(x)$ with minima is $-3$ at $x= 0$ and maxima is $3$ at $x = -2$ has graph as a conic curve.
How many critical point of $y = g(x) = 3^{\displaystyle{2f(x) + 4x -3}} - 2^{\displaystyle{-f(x) - 2x+3}}$ ?
Firsts step is derivative?
$g'(x) = [2f'(x) + 4]3^{\displaystyle{2f(x) + 4x -3}}.\ln(2f(x) + 4x +3)+ [f'(x) - 2]2^{\displaystyle{-f(x) - 2x+3}}.\ln(-f(x) - 2x +3)$
Then how to find solutions from $g'(x) = 0$ and given information?
I am confused, please help me!!!
You had the derivatives wrong, recall:
$$ a^{t(x)} = y \Rightarrow y' = a^{t(x)} \cdot t'(x) \cdot \ln(a) ~~ a \in \mathbb{R^+} $$
And so:
$$ g'(x) = 3^{2f(x)+4x-3} ( 2f'(x) + 4) \ln(3) - 2^{-f(x) -2x+3} (-f'(x) -2) \ln(2) \\ g'(x) = 3^{2f(x)+4x-3} (f'(x) + 2) 2\ln(3) + 2^{-f(x) -2x+3} (f'(x)+ 2) \ln(2) \\ (f'(x) +2) \left [ 3^{2f(x)+4x-3} 2 \ln(3) + 2^{-f(x) -2x+3} \ln(2) \right ] = 0$$
Now you can see how many possible values you have, because $$3^{2f(x)+4x-3} 2 \ln(3) + 2^{-f(x) -2x+3} \ln(2) > 0 $$ For all real $x$.