While doing optimisation problems with calculus I often find myself needing to optimise expressions such as \( \sqrt{\sin x} \), though often more complicated. Using the chain rule on the square root is time-consuming, and it seems logical that as a monotonically increasing function, the square root shouldn't affect the x-coordinate of the local minima and maxima. Thus, I give you…

Theorem

If \( g(f(x)) \) is a composite function, and \( g \) is a monotonic function, and is differentiable \( g \) in some domain;
Then the x-coordinates of the local minima and maxima of \( g(f(x)) \) in that domain are the same as the those of \( f(x) \).

Proof

Using the chain rule to find the derivative of \( g(f(x)) \) with respect to \( x \):
\[ \begin{aligned}
\frac{\mathrm{d}g}{\mathrm{d}x} &= \frac{\mathrm{d}g}{\mathrm{d}f} \cdot \frac{\mathrm{d}f}{\mathrm{d}x}
\end{aligned} \]
By Fermat's theorem, each local minimum and maximum of \( g(f(x)) \) must be at a critical point \( x_{\text{critical}} \); where \( \frac{\mathrm{d}g}{\mathrm{d}x} \) is equal to 0 or undefined:
\[ \begin{aligned}
\left(\frac{\mathrm{d}g}{\mathrm{d}x}\right)_{x = x_{\text{critical}}} &= 0\text{ or undefined} \\
\left(\frac{\mathrm{d}g}{\mathrm{d}f}\right)_{x = x_{\text{critical}}} \cdot \left(\frac{\mathrm{d}f}{\mathrm{d}x}\right)_{x = x_{\text{critical}}} &= 0\text{ or undefined}
\end{aligned} \]
By the zero-product property (null factor law),
\[ \begin{aligned}
\left(\frac{\mathrm{d}g}{\mathrm{d}f}\right)_{x = x_{\text{critical}}} &= 0\text{ or undefined} \\
&\text{and/or} \\ \left(\frac{\mathrm{d}f}{\mathrm{d}x}\right)_{x = x_{\text{critical}}} &= 0\text{ or undefined}
\end{aligned} \]
Since \( g \) is a monotonic function, \( \frac{\mathrm{d}g}{\mathrm{d}f} \) is never equal to 0. Additionally, since the derivative of \( g \) is always defined in the relevant domain, \( \frac{\mathrm{d}g}{\mathrm{d}f} \) is never undefined.
Therefore,
\[ \begin{aligned}
\left(\frac{\mathrm{d}f}{\mathrm{d}x}\right)_{x = x_{\text{critical}}} &= 0\text{ or undefined}
\end{aligned} \]
Therefore, each critical point on \( g(f(x)) \) is at the same x-coordinate as that on \( f(x) \), and by extension, each local minimum and maximum of \( g(f(x)) \) is at the same x-coordinate as that of \( f(x) \).

Q.E.D.

Disclaimer

I am not a mathematician. As far as I can tell, the above is not any ‘official’ theorem. There are probably some small errors in terminology and corner cases that I forgot to account for. Exercise common sense when applying the above.