Determine all $x \in \mathbb{R}$ so that $2^x = x^2+1$

Question

I am studying for my exam in two weeks and currently going over an old exam where I found the following task:

Determine all $x \in \mathbb{R}$ so that $2^x = x^2+1$. The hint is: Determine the first and second derivate of $f(x)=2^x-1-x^2$ and calculate $f(x)$ for $x \in {0, 1, , 2, 3, 4, 5}$. Then use Rolle's Theorem: https://en.wikipedia.org/wiki/Rolle%27s_theorem

So I did that:

$f(x) = 2^x-1-x^2$

$f'(x) = 2^xln(2)-2x$

$f''(x) = 2^xln^2(2)-2$

and $f(0)=0$, $f(1)=0$, $f(2)=-1$, $f(3)=-2$, $f(4)=-1$, $f(5)=6$

But how does that help me finding all $x \in \mathbb{R}$ so that $2^x = x^2+1$?

Answer 1

It may help to go as far as the third derivative. Since $f'''(x) = 2^x (\ln 2)^3 > 0$ for all $x$ you have that $f''$ is increasing. Thus $f''$ has at most one zero.

It's not hard at all to see that $f''$ has a zero at $x = \log_2 \left( \dfrac{2}{(\ln 2)^2}\right)$ but that isn't even really necessary to know.

The derivative $g'$ of any differentiable function $g$ has at least one zero in between any two distinct zeros of $g$. Turning that around, if $g'$ has at most one zero then $g$ has at most two zeros.

Since $f''$ has at most one zero, $f'$ has at most two and in turn $f$ has at most three.

Now follow the hint: $f(0) = 0$, $f(1) = 0$, $f(4) = -1$, and $f(5) = 6$.

You have found already two zeros of $f$. The intermediate value theorem gives you a third zero in between $4$ and $5$. By the remarks above there are no more zeros.

Answer 2

Let $f(x)=2^x-x^2-1$.

Thus, $$f''(x)=2^x\ln^22-1<0$$ for all $-1<x<2$, which says that on $[-1,2]$ our equation has two roots maximum.

But we saw that $0$ and $1$ they are roots.

The third root we can get by the continuous of $f$.

Answer 3

Note that no roots exist for $x < 0$ as in that interval, $2^x < 1$ but $x^2 + 1 > 1$. So analyse for $x \ge 0$. Now take $f(x) = 2^x -x^2-1$ and $f'(x) = 2^x \ln(2)-2x$ and $f''(x) = 2^x (\ln(2))^2 -2$.

Evidently $f''(x)$ has only one root since it is strictly increasing ($f'''(x) > 0$ ). Thus it is easy to follow from here that $f'(x)$ has two roots and $f(x)$ has three roots for $x \ge 0$ using rolle's.