4.6, 4.7 - Stationary points and the second derivative

4.6, 4.7 - Stationary points and the second derivative

Locate a curve's stationary points and decide whether each is a maximum, a minimum or neither. You will use the second derivative, which measures how the gradient changes, and check gradient signs when that test cannot decide.

Locating stationary points

Does a horizontal tangent always mark a maximum or a minimum? The curve y=x3y=x^3 has a horizontal tangent at the origin, yet it keeps rising through that point. We will explain this apparent exception after learning to locate every stationary point reliably.

You already know

The gradient function is dydx\frac{\mathrm{d}y}{\mathrm{d}x}. A positive gradient means the curve rises as you move right, and a negative gradient means it falls. For integer n≠0n\ne0, use ddx(kxn)=knxn−1\frac{\mathrm{d}}{\mathrm{d}x}(kx^n)=knx^{n-1}; constants differentiate to zero.

A horizontal tangent has gradient zero. That gives an equation for the input xx, but not the point's height. After solving it, substitute into the original curve equation for yy.

Stationary point

A stationary point is a point on a differentiable curve where dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0. Its tangent is horizontal. A turning point is a local maximum or minimum, where the curve changes from rising to falling or from falling to rising. A stationary point need not be a turning point.

The word “local” compares a point with nearby points on the curve. A local maximum need not be the highest point on the entire graph.

Stationary-point coordinates

At an interior point where the derivative exists,

dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0

locates stationary inputs. Substitute each allowed input into the original expression for yy to obtain its coordinates. This condition alone does not establish a maximum or minimum.
Not on the formulae sheet: learn it.

Keep the gradient equation and the height calculation on separate lines.

Worked example 1: two stationary points

Question. Work out the coordinates of the stationary points of y=x3−3xy=x^3-3x.

Plan. Differentiate, solve the zero-gradient equation and substitute both inputs into the curve.

dydx=3x2−3differentiate3x2−3=0horizontal tangentx2=1⟹x=−1 or 1both square rootsy(−1)=(−1)3−3(−1)=2use the curve equationy(1)=1−3=−2use the other input\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-3 &&\text{differentiate}\\ 3x^2-3&=0 &&\text{horizontal tangent}\\ x^2&=1\quad\Longrightarrow\quad x=-1\text{ or }1 &&\text{both square roots}\\ y(-1)&=(-1)^3-3(-1)=2 &&\text{use the curve equation}\\ y(1)&=1-3=-2 &&\text{use the other input} \end{aligned}

Why this step? Solving 3x2−3=03x^2-3=0 gives inputs. It does not mean the outputs are zero.

Answer. The stationary points are (−1,2)(-1,2) and (1,−2)(1,-2).

Check. At both inputs, 3x2−3=03x^2-3=0, so both tangents are horizontal. Each coordinate pair also satisfies the original equation.

Worked example 2 adds only a constant, 44, to the curve. Its heights change, but its gradients do not.

Worked example 2: shifting the curve upwards

Question. Work out the coordinates of the stationary points of y=x3−3x+4y=x^3-3x+4.

Plan. Solve the same gradient equation, then use the new curve for the heights.

dydx=3x2−3the constant differentiates to zero3x2−3=0⟹x=−1 or 1stationary inputsy(−1)=−1+3+4=6new heighty(1)=1−3+4=2new height\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-3 &&\text{the constant differentiates to zero}\\ 3x^2-3&=0\quad\Longrightarrow\quad x=-1\text{ or }1 &&\text{stationary inputs}\\ y(-1)&=-1+3+4=6 &&\text{new height}\\ y(1)&=1-3+4=2 &&\text{new height} \end{aligned}

Why this step? Moving a curve vertically preserves its tangent directions. The same stationary inputs now have outputs 44 greater.

Answer. The stationary points are (−1,6)(-1,6) and (1,2)(1,2).

Check. Adding 44 to the earlier heights 22 and −2-2 gives 66 and 22, as expected.

The derivative tells you where to look; the original function tells you where the point sits.

Watch out

Wrong: dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0, so the stationary point has y=0y=0. Right: the gradient is zero. Calculate yy separately; neither point in Worked example 2 lies on the xx-axis.

Use new coefficients in the next stationary-point calculation.

Measuring how the gradient changes

Differentiate the gradient function to measure how quickly the gradient itself changes as xx increases.

For y=x2−4xy=x^2-4x, the gradient is 2x−42x-4. At x=1x=1, the gradient is −2-2, so the curve falls. Yet the gradient function is increasing: its gradient is 22. A rising gradient and a rising curve are different statements.

Second derivative

The second derivative, written d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}, is the derivative of the gradient function. It measures the rate of change of dydx\frac{\mathrm{d}y}{\mathrm{d}x} with respect to xx. It is not the square of the first derivative.

A positive second derivative means the gradient is increasing locally; a negative one means the gradient is decreasing locally. An increasing gradient can still be negative, as a descending curve becomes less steep.

Differentiating twice

Where the relevant derivatives exist,

d2ydx2=ddx(dydx)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)

For the integer powers used here, apply the power rule again to every term of the first derivative.
Not on the formulae sheet: learn it.

Negative powers follow exactly the same rule. Rewrite a reciprocal as a power before differentiating.

Worked example 3: differentiating a reciprocal twice

Question. For y=x+9xy=x+\frac9x, where x>0x>0, work out d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} and interpret its sign.

Plan. Write 9/x9/x as 9x−19x^{-1}, then differentiate twice.

y=x+9x−1integer-power formdydx=1−9x−2first derivatived2ydx2=(−9)(−2)x−3=18x−3differentiate the gradient\begin{aligned} y&=x+9x^{-1} &&\text{integer-power form}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=1-9x^{-2} &&\text{first derivative}\\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}&=(-9)(-2)x^{-3}=18x^{-3} &&\text{differentiate the gradient} \end{aligned}

Why this step? The first derivative's coefficient is already negative. Multiplying it by the exponent −2-2 gives a positive coefficient.

Answer. d2ydx2=18x3>0\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{18}{x^3}>0 for x>0x>0, so the gradient is increasing throughout the domain.

Check. At x=3x=3, the second derivative is 18/27=2/318/27=2/3, positive. The first derivative there is 00, showing that the two derivatives have different roles.

Change only the plus sign before the reciprocal to minus. Track the sign through both differentiations.

Worked example 4: reversing the reciprocal's sign

Question. For y=x−9xy=x-\frac9x, where x>0x>0, work out d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} and interpret its sign.

Plan. Differentiate x−9x−1x-9x^{-1} twice.

y=x−9x−1rewrite the reciprocaldydx=1+9x−2negative times negatived2ydx2=9(−2)x−3=−18x−3differentiate again\begin{aligned} y&=x-9x^{-1} &&\text{rewrite the reciprocal}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=1+9x^{-2} &&\text{negative times negative}\\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}&=9(-2)x^{-3}=-18x^{-3} &&\text{differentiate again} \end{aligned}

Why this step? You differentiate the current expression, 1+9x−21+9x^{-2}, rather than returning to the original power −1-1.

Answer. d2ydx2=−18x3<0\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{18}{x^3}<0 for x>0x>0, so the gradient is decreasing throughout the domain.

Check. At x=3x=3, the second derivative is −2/3-2/3. The first derivative is 22: the curve is rising even though its gradient is decreasing.

The sign of the second derivative describes the change in slope. It does not by itself tell you whether the curve is above the axis or rising.

Watch out

Wrong: differentiating −9x−2-9x^{-2} gives −18x−3-18x^{-3}. Right: (−9)(−2)x−3=18x−3(-9)(-2)x^{-3}=18x^{-3}.

Apply the same two differentiations to another reciprocal expression.

Classifying maxima and minima

At a stationary point, the sign of the second derivative usually tells you which way the curve turns.

At a minimum, the gradient changes from negative to positive: it is increasing through zero. At a maximum, it changes from positive to negative: it is decreasing through zero. This explains the signs in the test.

Second-derivative test

For a stationary point at x=ax=a on a curve with a continuous second derivative near aa,

d2ydx2∣x=a>0⟹local minimumd2ydx2∣x=a<0⟹local maximum\begin{aligned} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg|_{x=a}>0&\quad\Longrightarrow\quad\text{local minimum}\\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg|_{x=a}<0&\quad\Longrightarrow\quad\text{local maximum} \end{aligned}

If the second derivative is zero, the test is inconclusive. Check gradient signs on both sides instead.
Not on the formulae sheet: learn it.

Always establish stationarity first. A positive second derivative at an arbitrary point does not make that point a minimum.

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