Stationary Points

Finding Stationary Points

To find a point where the gradient is $0$, we need to set $dfrac{dy}{dx} = f'(x) = 0$.

So, let’s say we’ve got a nice, easy function: $textcolor{blue}{y = x^2 – 1}$.

Well, we know that the gradient is given by $dfrac{dy}{dx} = 2x$.

Setting $textcolor{limegreen}{dfrac{dy}{dx} = 2x = 0}$ gives $x = 0$, for which $y = 0^2 – 1 = -1$.

We conclude that $textcolor{blue}{y = x^2 – 1}$ has a stationary point at $(0, -1)$.

Finding Stationary Points of Complicated Functions

Let $textcolor{blue}{y = sin x + cos x}$ for $x$ measured in radians. How might we go about finding the stationary points (in the range
$0 leq x leq 2pi$)?

$textcolor{blue}{y = sin x + cos x}$ gives $dfrac{dy}{dx} = cos x – sin x$.

Setting $textcolor{limegreen}{dfrac{dy}{dx} = 0}$, we require $cos x = sin x$, or, $tan x = 1$.

This gives $x = tan ^{-1}(1) = dfrac{pi}{4}, dfrac{5pi}{4}$.

Therefore, for $0 leq x leq 2pi$, we have the stationary points $left( dfrac{pi}{4}, sqrt{2}right) , left( dfrac{5pi}{4}, -sqrt{2}right)$.

Finding the Nature of Stationary Points

Once we’ve found our stationary points, we need to find out whether they are a maximum, minimum, or a stationary point of inflection.

We do this by differentiating our derivative again.

So, for example, take our first example of $y = x^2 – 1$.

We said that $dfrac{dy}{dx} = 2x$.

So $dfrac{d^{2}y}{dx^2} = 2$.

Since $dfrac{d^{2}y}{dx^2} = 2 > 0$, we can conclude that this is a minimum point. You probably figured that out from the graph, though.

What about the second example, $y = sin x + cos x$?

Well, we know that $dfrac{dy}{dx} = cos x – sin x$.

By extension, $dfrac{d^{2}y}{dx^2} = -sin x – cos x$. We now need to test both of our values of $x$ for the corresponding value of $dfrac{d^{2}y}{dx^2}$.

So, for $x = dfrac{pi}{4}$, $dfrac{d^{2}y}{dx^2} = -sqrt{2} < 0$, so it is a maximum point.

For $x = dfrac{5pi}{4}$, $dfrac{d^{2}y}{dx^2} = sqrt{2} > 0$, so it is a minimum point.

Sketching Graphs from Information about Functions

Say we have a complex function with multiple terms, i.e. $textcolor{blue}{f(x) = 1 + x – 2cos x}$ where $-2pi leq x leq 2pi$ is measured in radians.

Then $f'(x) = 1 + 2sin x$ and $f''(x) = 2cos x$.

Therefore, any stationary points will occur when $sin x = dfrac{-1}{2}$, so $x = dfrac{-5pi}{6}, dfrac{-pi}{6}, dfrac{7pi}{6}, dfrac{11pi}{6}$.

For $x = dfrac{-5pi}{6}$, $f(x) = 0.114$ and $f''(x) = -sqrt{3}$, which is a maximum.

For $x = dfrac{-pi}{6}$, $f(x) = -1.256$ and $f''(x) = sqrt{3}$, which is a minimum.

For $x = dfrac{7pi}{6}$, $f(x) = 6.397$ and $f''(x) = -sqrt{3}$, which is a maximum.

For $x = dfrac{11pi}{6}$, $f(x) = 5.028$ and $f''(x) = sqrt{3}$, which is a minimum.

We can sketch these parts onto a graph to give us a rough form.

We can be helped further in sketching by considering the behaviour of the graph as $x$ gets really big or really small. We know that the graph roughly follows $y=x+1$, so as $x$ gets really big so does $y$, and as $x$ gets really small so does $y$. With this, we can finish our graph.