Mathematics 9709 · Pure Mathematics 1

Stationary points and the second-derivative test

A stationary point is where a curve is flat for a moment (its tangent is horizontal). Think of the top of a hill or the bottom of a valley. It is where the gradient is zero.
Use it to find turning points and say which is which. This is one of the most common long-question parts in P1.

Find them: solve $\dfrac{dy}{dx}=0$ . Then put each $x$ back into the original curve to get the $y$ -coordinate.

Classify them with the second derivative: $\tfrac{d^2y}{dx^2}<0$ is a maximum (frowning), and $\tfrac{d^2y}{dx^2}>0$ is a minimum (smiling).

Finding them: set the gradient to zero

A stationary point is where $\dfrac{dy}{dx}=0$ . Differentiate, set it to zero, then solve for $x$ .
Then put each $x$ back into the curve (not the derivative) to get the $y$ -coordinate.

At a maximum and a minimum the tangent is horizontal: dy/dx = 0.

Drag the point below along $y=x^3-3x$ . The blue tangent tilts and the gradient number changes.
The moment the gradient is zero you are at a stationary point. The turning point lights up: gold for the maximum at $(-1,2)$ , green for the minimum at $(1,-2)$ .

drag P

at x = -1.70·gradient dy/dx = +5.7·uphill

Classifying them: the second derivative

The exam also wants the nature: is it a maximum or a minimum?

Differentiate a second time to get $\dfrac{d^2y}{dx^2}$ , then check if it is positive or negative at each stationary point:

$\dfrac{d^2y}{dx^2}<0\;\Rightarrow\;\text{maximum},\qquad \dfrac{d^2y}{dx^2}>0\;\Rightarrow\;\text{minimum}$

Here is a way to remember which is which. At a maximum the curve frowns (it bends downwards). The gradient is falling as you pass through, so $\tfrac{d^2y}{dx^2}<0$ .
At a minimum the curve smiles (it bends upwards). The gradient rises, so $\tfrac{d^2y}{dx^2}>0$ .

Worked example

Warm-up: find and classify the stationary point of $y = x^2 - 4x$

Differentiate and set it to zero: $\dfrac{dy}{dx}=2x-4=0\Rightarrow x=2$ . The point is $(2,-4)$ .
Second derivative: $\dfrac{d^2y}{dx^2}=2>0$ , so it is a minimum.

Answer

\text{min }(2,-4)

Worked example

Real question: find and classify the stationary points of $y = 54x - (2x-7)^3$

9709/12 May/June 2020 Q10(b),(c). The chain-rule derivative from §7.2 is used again here.

From part (a), $\dfrac{dy}{dx}=54-6(2x-7)^2$ . Set it to zero: $6(2x-7)^2=54\Rightarrow(2x-7)^2=9$ .
Square-root both ways: $2x-7=\pm 3$ , so $x=5$ or $x=2$ . Back into the curve: points $(5,243)$ and $(2,135)$ .
Second derivative $\dfrac{d^2y}{dx^2}=-24(2x-7)$ . At $x=5$ : $-24(3)=-72<0$ (maximum). At $x=2$ : $-24(-3)=72>0$ (minimum).

Answer

\text{max }(5,243),\quad\text{min }(2,135)

If the second derivative comes out 0

If $\tfrac{d^2y}{dx^2}=0$ at a stationary point, this test tells you nothing. It could be a max, a min, or a point of inflexion. Instead, check the sign of $\tfrac{dy}{dx}$ just before and just after the point. If the gradient goes $+\,0\,-$ it is a maximum. If it goes $-\,0\,+$ it is a minimum. If the sign is the same on both sides (for example $+\,0\,+$ ) it is a stationary point of inflexion.

What the exam asks

Often the curve has an unknown constant. The phrase “has a stationary point at $x=2$ ” fixes it: set $\tfrac{dy}{dx}=0$ at that point and find the constant.
A common next part: if there is only one stationary point, its $y$ -value is the edge of the function's range. (9709/12 Oct/Nov 2021 Q10)

Where the marks go

Common mistake

Giving the coordinates but not the nature loses the classification marks. Always test with

\tfrac{d^2y}{dx^2}

(or a sign check). Write the answer in words too. (9709/12 May/June 2020 Q10)

Common mistake

Do not get the test the wrong way round.

\tfrac{d^2y}{dx^2}>0

is a minimum (the smile).

<0

is a maximum. And to get the

y

-value, put

x

into the original curve, not the derivative.

Now you try

The function f is $f(x)=x^2+\dfrac{k}{x}+2$ for $x>0$ . It has a stationary point at $x=2$ . Find $k$ , then say if the stationary point is a maximum or a minimum. (9709/12 Oct/Nov 2021 Q10)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q10)

Write $f(x)=x^2+kx^{-1}+2$ , so $f'(x)=2x-kx^{-2}$ .

Stationary at $x=2$ : $2(2)-\tfrac{k}{4}=0\Rightarrow k=16$ .

Then $f''(x)=2+2kx^{-3}=2+32x^{-3}$ , and at $x=2$ , $f''(2)=2+4=6>0$ .

Answer

k=16,\quad\text{minimum point}

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Stationary points and the second-derivative test

A stationary point is where a curve is flat for a moment (its tangent is horizontal). Think of the top of a hill or the bottom of a valley. It is where the gradient is zero.
Use it to find turning points and say which is which. This is one of the most common long-question parts in P1.

Find them: solve $\dfrac{dy}{dx}=0$ . Then put each $x$ back into the original curve to get the $y$ -coordinate.

Classify them with the second derivative: $\tfrac{d^2y}{dx^2}<0$ is a maximum (frowning), and $\tfrac{d^2y}{dx^2}>0$ is a minimum (smiling).

Finding them: set the gradient to zero

A stationary point is where $\dfrac{dy}{dx}=0$ . Differentiate, set it to zero, then solve for $x$ .
Then put each $x$ back into the curve (not the derivative) to get the $y$ -coordinate.

At a maximum and a minimum the tangent is horizontal: dy/dx = 0.

Drag the point below along $y=x^3-3x$ . The blue tangent tilts and the gradient number changes.
The moment the gradient is zero you are at a stationary point. The turning point lights up: gold for the maximum at $(-1,2)$ , green for the minimum at $(1,-2)$ .

drag P

at x = -1.70·gradient dy/dx = +5.7·uphill

Classifying them: the second derivative

The exam also wants the nature: is it a maximum or a minimum?

Differentiate a second time to get $\dfrac{d^2y}{dx^2}$ , then check if it is positive or negative at each stationary point:

$\dfrac{d^2y}{dx^2}<0\;\Rightarrow\;\text{maximum},\qquad \dfrac{d^2y}{dx^2}>0\;\Rightarrow\;\text{minimum}$

Here is a way to remember which is which. At a maximum the curve frowns (it bends downwards). The gradient is falling as you pass through, so $\tfrac{d^2y}{dx^2}<0$ .
At a minimum the curve smiles (it bends upwards). The gradient rises, so $\tfrac{d^2y}{dx^2}>0$ .

Worked example

Warm-up: find and classify the stationary point of $y = x^2 - 4x$

Differentiate and set it to zero: $\dfrac{dy}{dx}=2x-4=0\Rightarrow x=2$ . The point is $(2,-4)$ .
Second derivative: $\dfrac{d^2y}{dx^2}=2>0$ , so it is a minimum.

Answer

\text{min }(2,-4)

Worked example

Real question: find and classify the stationary points of $y = 54x - (2x-7)^3$

9709/12 May/June 2020 Q10(b),(c). The chain-rule derivative from §7.2 is used again here.

From part (a), $\dfrac{dy}{dx}=54-6(2x-7)^2$ . Set it to zero: $6(2x-7)^2=54\Rightarrow(2x-7)^2=9$ .
Square-root both ways: $2x-7=\pm 3$ , so $x=5$ or $x=2$ . Back into the curve: points $(5,243)$ and $(2,135)$ .
Second derivative $\dfrac{d^2y}{dx^2}=-24(2x-7)$ . At $x=5$ : $-24(3)=-72<0$ (maximum). At $x=2$ : $-24(-3)=72>0$ (minimum).

Answer

\text{max }(5,243),\quad\text{min }(2,135)

If the second derivative comes out 0

What the exam asks

Often the curve has an unknown constant. The phrase “has a stationary point at $x=2$ ” fixes it: set $\tfrac{dy}{dx}=0$ at that point and find the constant.
A common next part: if there is only one stationary point, its $y$ -value is the edge of the function's range. (9709/12 Oct/Nov 2021 Q10)

Where the marks go

Common mistake

Giving the coordinates but not the nature loses the classification marks. Always test with

\tfrac{d^2y}{dx^2}

(or a sign check). Write the answer in words too. (9709/12 May/June 2020 Q10)

Common mistake

Do not get the test the wrong way round.

\tfrac{d^2y}{dx^2}>0

is a minimum (the smile).

<0

is a maximum. And to get the

y

-value, put

x

into the original curve, not the derivative.

Now you try

The function f is $f(x)=x^2+\dfrac{k}{x}+2$ for $x>0$ . It has a stationary point at $x=2$ . Find $k$ , then say if the stationary point is a maximum or a minimum. (9709/12 Oct/Nov 2021 Q10)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q10)

Write $f(x)=x^2+kx^{-1}+2$ , so $f'(x)=2x-kx^{-2}$ .

Stationary at $x=2$ : $2(2)-\tfrac{k}{4}=0\Rightarrow k=16$ .

Then $f''(x)=2+2kx^{-3}=2+32x^{-3}$ , and at $x=2$ , $f''(2)=2+4=6>0$ .

Answer

k=16,\quad\text{minimum point}