Range of the function $f(x) = 4^x + 4^{-x} + 2^x + 2^{-x} + 3$

Question

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?

Answer 1

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:

• $f_1(x) = 2^x$ (this gives us $u$ in terms of $x$);
• $f_2(u) = u + 1/u$ (this gives us $t$ in terms of $u$);
• $f_3(t) = t^2+t+1$ (this gives us $f(x)$ in terms of $t$).

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:

• $f_1$ sends the real line $\mathbb R$ to the set $(0,\infty)$, because $2^x > 0$ for all $x$.
• Though $f_2$ sends the real line $\mathbb R$ to the set $[2,\infty)$, because $u + 1/u \ge 2$ for all $u$, what we really should ask is what it does to $(0,\infty)$: the range of $f_1$. In this case, that's still $[2,\infty)$: the minimum of $u+1/u=2$ is achieved at $u=1$ and from there, $u+1/u$ keeps increasing as we increase $u$.
• Though $f_3$ sends the real line $\mathbb R$ to the set $[3/4,\infty)$, as you've found, it sends $[2,\infty)$ to $[7,\infty)$. We have $f_3(2)=7$ and from there, the quadratic increases as we increase $t$.

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.

Answer 2

Notice that in order for $t^2+t+1$ to be $\frac{3}{4}$, $t$ should be $-\frac{1}{2}$. But $t$ cannot be $-\frac{1}{2}$, because the equation $u+\frac{1}{u}=t$ doesn't allow it. In fact, $t$ cannot be anywhere below $2$ or even $2$ itself—the $u+\frac{1}{u}=t$ equation becomes invalid for such $t$.

So $t$ has some responsibilities towards $u$: it must be greater than $2$. Notice that putting $t=2$ in $t^2+t+1$, yields $7$. Which is what we expect for the lower bound of our range.