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.
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http://math.stackexchange.com/questions/1922417/gre-subject-test-in-mathematics-where-can-i-find-related-past-papers-solution – BCLC Oct 26 '16 at 15:15
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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 $x = 0$.
sTertooy
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