Stationary Points | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

A stationary point has zero gradient

At a stationary point the tangent is horizontal, so $f'(x)=0$ . Solving this gives the $x$ -coordinate. Substitute into the original function to find the height $y$ .

Worked example

Find the stationary points of y = x³ − 3x.

3x^2-3=0\ \Longrightarrow\ x=\pm1

$f(-1)=2$ and $f(1)=-2$ , so the points are $(-1,2)$ and $(1,-2)$ . Substitution into the derivative would only give zero again.

Decide whether it is a maximum or a minimum

A local maximum is higher than nearby curve points; a local minimum is lower. “Local” does not mean greatest or least on the whole graph.

Read left to right. A positive-to-negative gradient gives rise then fall: a maximum. A negative-to-positive gradient gives fall then rise: a minimum. This is the first derivative test.

Both tangents are horizontal. The gradient signs on either side distinguish maximum from minimum.

The second derivative measures how the gradient changes. At a stationary point, $f''(a)<0$ means the gradient is decreasing through zero: a maximum. If $f''(a)>0$ , it is increasing through zero: a minimum.

Worked example

Classify the two points just found.

$f''(x)=6x$ . At $x=-1$ , $f''(-1)=-6<0$ : maximum at $(-1,2)$ . At $x=1$ , $f''(1)=6>0$ : minimum at $(1,-2)$ .

Common mistake

Check that the first derivative is zero before using the second derivative to classify a stationary point.

Your turn9709/12 May/June 2020 Q10(b–c) [5 marks]

Find the stationary points of $y=54x-(2x-7)^3$ and determine their nature.

Show worked answer

\begin{aligned}y'&=54-6(2x-7)^2\\y'=0&\Rightarrow (2x-7)^2=9\\&\Rightarrow 2x-7=\pm3\\&\Rightarrow x=2\text{ or }5\end{aligned}

Substitute into the curve: $y(2)=135$ , $y(5)=243$ . Now $y''=-24(2x-7)$ gives $72>0$ at 2 and $-72<0$ at 5.

Answer

\begin{gathered}(2,135)\text{ minimum}\\(5,243)\text{ maximum}\end{gathered}

If the second derivative is zero

The second derivative test gives no decision when $f''(a)=0$ . Use the sign of the first derivative immediately on either side, without crossing another zero or an excluded value.

Worked example

y = (x − 1)⁴ has a stationary point at (1, 0).

$y'=4(x-1)^3$ and $y''=12(x-1)^2$ . Although $y''(1)=0$ , $y'$ changes from negative to positive at 1. The point is a minimum.

For $y=x^3$ , $y'=3x^2$ is positive on both sides of zero. The stationary point is neither a maximum nor a minimum. A horizontal tangent need not be a turn.

Your turnPractice

Classify the stationary point of $y=-(x+2)^4$ . Explain why the second derivative alone is insufficient.

Show worked answer

At $x=-2$ , $y=0$ and $y''=0$ . But $y'=-4(x+2)^3$ changes from positive to negative, so $(-2,0)$ is a maximum.

A stationary point has zero gradient

At a stationary point the tangent is horizontal, so $f'(x)=0$ . Solving this gives the $x$ -coordinate. Substitute into the original function to find the height $y$ .

Worked example

Find the stationary points of y = x³ − 3x.

3x^2-3=0\ \Longrightarrow\ x=\pm1

$f(-1)=2$ and $f(1)=-2$ , so the points are $(-1,2)$ and $(1,-2)$ . Substitution into the derivative would only give zero again.

Decide whether it is a maximum or a minimum

A local maximum is higher than nearby curve points; a local minimum is lower. “Local” does not mean greatest or least on the whole graph.

Read left to right. A positive-to-negative gradient gives rise then fall: a maximum. A negative-to-positive gradient gives fall then rise: a minimum. This is the first derivative test.

Both tangents are horizontal. The gradient signs on either side distinguish maximum from minimum.

Worked example

Classify the two points just found.

$f''(x)=6x$ . At $x=-1$ , $f''(-1)=-6<0$ : maximum at $(-1,2)$ . At $x=1$ , $f''(1)=6>0$ : minimum at $(1,-2)$ .

Common mistake

Check that the first derivative is zero before using the second derivative to classify a stationary point.

Your turn9709/12 May/June 2020 Q10(b–c) [5 marks]

Find the stationary points of $y=54x-(2x-7)^3$ and determine their nature.

Show worked answer

\begin{aligned}y'&=54-6(2x-7)^2\\y'=0&\Rightarrow (2x-7)^2=9\\&\Rightarrow 2x-7=\pm3\\&\Rightarrow x=2\text{ or }5\end{aligned}

Substitute into the curve: $y(2)=135$ , $y(5)=243$ . Now $y''=-24(2x-7)$ gives $72>0$ at 2 and $-72<0$ at 5.

Answer

\begin{gathered}(2,135)\text{ minimum}\\(5,243)\text{ maximum}\end{gathered}

If the second derivative is zero

The second derivative test gives no decision when $f''(a)=0$ . Use the sign of the first derivative immediately on either side, without crossing another zero or an excluded value.

Worked example

y = (x − 1)⁴ has a stationary point at (1, 0).

$y'=4(x-1)^3$ and $y''=12(x-1)^2$ . Although $y''(1)=0$ , $y'$ changes from negative to positive at 1. The point is a minimum.

For $y=x^3$ , $y'=3x^2$ is positive on both sides of zero. The stationary point is neither a maximum nor a minimum. A horizontal tangent need not be a turn.

Your turnPractice

Classify the stationary point of $y=-(x+2)^4$ . Explain why the second derivative alone is insufficient.

Show worked answer

At $x=-2$ , $y=0$ and $y''=0$ . But $y'=-4(x+2)^3$ changes from positive to negative, so $(-2,0)$ is a maximum.