The number of real roots of $e^x = x^2 $ .

Question

The number of real roots of $e^x = x^2 $ .

According to me, it should me zero. But the answer says its one.

How to proceed.

Answer 1

Let $f(x) = e^x - x^2$, then $f'(x) = e^x - 2x$ and $f''(x) = e^x - 2$. $f''(x) = 0$ at $x = \ln(2)$ , and so $f'(x) \geq f'(\ln(2)) = 2 - 2 \ln(2)$. Since $2 < e$ we have $f'(x) > 0$ everywhere.

Thus the function either has 0 roots or 1 root. Since $\lim_{x\to -\infty} f(x) = - \infty$ and $\lim_{x \to \infty} f(x) = \infty$ we see it has one root.

Answer 2

$(1)\enspace x>0$ : $\enspace x^{\frac{1}{x}}=\sqrt{e}>\max(x^{\frac{1}{x}})=e^{\frac{1}{e}}\enspace $ => no solution

$(2)\enspace y:=-x$ with $y>0$ : $\enspace y^{\frac{1}{y}}=\frac{1}{\sqrt{e}}<1\enspace $ => one solution

With $(1)+(2)$ follows: one solution