# Optimisation of a composite function involving a differentiable monotonic component

# 0

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} \]

\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} \]

\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} \]

\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} \]

\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.