finding the number of all real zeros of $f(x)=\sec(x)-e^{-x^2}$

Question

How can we find the number of all real zeros of $f(x)=\sec(x)-e^{-x^2}$? This was an exam question of GRE subject. Also I am struggle with sketching the graph $f$ by hand.

Answer 1

Rather than drawing $f$, draw $y = \sec{x}$ and $y = e^{-x^2}$. The question asks for $x$ such that $f(x) = 0$, meaning those $x$ where $\sec(x) = e^{-x^2}$.

Since $\sec{x} = \frac{1}{\cos{x}}$ and $\cos{x} \in [-1,1]$, you know that $\sec{x} \in (-\infty,-1] \cup [1,\infty)$ in those $x$ where it is defined.

On the other hand, $e^{-x^2} \in (0,1]$ for all $x$.

Hence the only possible point of intersection is when both functions take the value of $1$, which happens when $x = 0$.