A stationary point of $y=\mathrm{f}(x)$ is a point where \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0

Question

A stationary point of $y=\mathrm{f}(x)$ is a point where $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$. Solving $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ only finds the candidates — a separate test decides whether each one is a maximum point, a minimum point or a stationary point of inflexion. $\small\text{Stationary Point, }\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ $\small\textcolor{#1E6FBA}{\textbf{MAXIMUM POINT}}$ $\small\textcolor{#1E6FBA}{\textbf{MINIMUM POINT}}$ $\small\textcolor{#1E6FBA}{\textbf{POINT OF INFLEXION}}$ $\small\begin{gathered}\textcolor{#1E6FBA}{\textbf{FIRST DERIVATIVE TEST}}\[12pt]\begin{array}{|c|c|c|c|}\hline x&a^{-}&a&a^{+}\[3pt]\hline\dfrac{\mathrm{d}y}{\mathrm{d}x}&\rlap{\text{Positive}}\hphantom{\text{Negative}}&0&\text{Negative}\[3pt]\hline\end{array}\end{gathered}$ $\small\begin{gathered}\textcolor{#1E6FBA}{\textbf{FIRST DERIVATIVE TEST}}\[12pt]\begin{array}{|c|c|c|c|}\hline x&a^{-}&a&a^{+}\[3pt]\hline\dfrac{\mathrm{d}y}{\mathrm{d}x}&\text{Negative}&0&\rlap{\text{Positive}}\hphantom{\text{Negative}}\[3pt]\hline\end{array}\end{gathered}$ $\small\begin{gathered}\textcolor{#1E6FBA}{\textbf{FIRST DERIVATIVE TEST}}\[12pt]\begin{array}{|c|c|c|c|}\hline x&a^{-}&a&a^{+}\[3pt]\hline\dfrac{\mathrm{d}y}{\mathrm{d}x}&\text{Negative}&0&\text{Negative}\[3pt]\hline\end{array}\[24pt]\begin{array}{|c|c|c|c|}\hline x&a^{-}&a&a^{+}\[3pt]\hline\dfrac{\mathrm{d}y}{\mathrm{d}x}&\rlap{\text{Positive}}\hphantom{\text{Negative}}&0&\rlap{\text{Positive}}\hphantom{\text{Negative}}\[3pt]\hline\end{array}\end{gathered}$ $\small\begin{gathered}\textcolor{#1E6FBA}{\textbf{SECOND DERIVATIVE TEST}}\[8pt]\mathrm{f}''(x)<0\[6pt]\text{As the gradient function is decreasing.}\end{gathered}$ $\small\begin{gathered}\textcolor{#1E6FBA}{\textbf{SECOND DERIVATIVE TEST}}\[8pt]\mathrm{f}''(x)>0\[6pt]\text{As the gradient function is increasing.}\end{gathered}$ $\small\begin{gathered}\textcolor{#1E6FBA}{\textbf{SECOND DERIVATIVE TEST}}\[8pt]\mathrm{f}''(x)=0\text{ and }\mathrm{f}''(x)\text{ changes signs.}\end{gathered}$ Gradient behaviour through $x=a$Stationary-point type $+$ to $-$ $-$ to $+$ No sign change

Accepted Answer

At a stationary point $\mathrm f'(a)=0$. Maximum: $\mathrm f'$ changes $+$ to $-$, or $\mathrm f''(a)0$. Stationary inflexion: $\mathrm f'$ keeps the same sign while concavity changes.

Gradient behaviour through \(x=a\)	Stationary-point type
\(+\) to \(-\)
\(-\) to \(+\)
No sign change

Stationary Points

Stationary Points

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