I tried to form a quadratic equation by taking $2^x = u$
$$u^2 + \frac{1}{u^2}+u + \frac{1}{u}+3$$
$$\left(u + \frac{1}{u}\right)^2+u + \frac{1}{u}+1$$
taking $u+1/u = t$
$$t^2+t+1$$
From this quadratic equation, range should be $[3/4, \infty)$. But this is wrong, correct answer is $[7, \infty)$ which I got using AM-GM inequality. My question is, why is the above quadratic method wrong?
Range of the function $f(x) = 4^x + 4^{-x} + 2^x + 2^{-x} + 3$
functionsquadratics
Best Answer
You should be clear on what you're doing when you're making the substitutions $u = 2^x$ and $t = u + 1/u$: you're writing your function as a composition of functions $$f(x) = f_3(f_2(f_1(x)))$$ where:
When the range is computed, we see look at each function in order from inside out, whereas you're only looking at the outermost function. So you should reason as follows:
In this case, looking carefully at $f_2$ turns out not to matter, but we should still do it. When we look carefully at $f_3$, it does change the range.