# Solution of exp equation

No. of Real solution of $2^x-x^2 = 1$

Solution (Given as):: Clearly $x = 0$ and $x = 1$ are solution of Given equation.

Formal; define $f(x) = 2^x - x^2 - 1$, so $f'(x) = 2^x\ln 2 - 2x$, $f''(x) = 2^x\ln^2 2 - 2$,

$f'''(x) = 2^x\ln^3 2 > 0$. Study the variation, in order to estimate the values of the roots and

critical points

Now I Did not Under How can we check Nature of $f(x)$ using $f^{'''}(x)$

Thanked