CA5, CA6 - Finding and classifying stationary points

CA5, CA6 - Finding and classifying stationary points

A horizontal tangent is a clue, not a verdict about a curve's shape. CA5 shows how to locate stationary points by differentiation; CA6 shows how to decide whether each is a maximum, a minimum or neither. You will give both coordinates and support every classification with a reason.

Finding stationary points

The graphs y=x3y=x^3 and y=x4y=x^4 both have gradient zero at the origin. Does each turn there? One keeps rising straight through the flat point; the other reaches its lowest height. We need a way to separate “flat” from “turning”.

You already know

The derivative gives a tangent's gradient. Differentiate a polynomial term by term; solve a quadratic by factorising where possible. Substitute into the original curve to find a height.

At a stationary point the gradient is zero, so first solve the derivative equation. The solutions locate possible changes of direction, but do not yet tell you what those changes are.

Stationary point

A stationary point is a point on a curve where dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0, so the tangent is horizontal. For example, the origin is stationary on y=x2y=x^2. The point (1,1)(1, 1) is not stationary on that curve, since its gradient is 22.

The condition f′(x)=0\mathrm{f}'(x)=0 for a stationary point is not on the formulae sheet: learn it. Notice that f(x)=0\mathrm{f}(x)=0 instead finds horizontal-axis intersections.

Worked example 1: two stationary points

Question. Determine the coordinates of the stationary points of y=x3−3x+1y=x^3-3x+1.

Plan. Differentiate, set the derivative equal to zero and find each corresponding height.

dydx=3x2−3gradient function3x2−3=0stationary conditionx2=1⇒x=−1 or 1solvey(−1)=−1+3+1=3original curvey(1)=1−3+1=−1original curve\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-3 && \text{gradient function}\\ 3x^2-3&=0 && \text{stationary condition}\\ x^2&=1\quad\Rightarrow\quad x=-1\text{ or }1 && \text{solve}\\ y(-1)&=-1+3+1=3 && \text{original curve}\\ y(1)&=1-3+1=-1 && \text{original curve} \end{aligned}

Why use the original formula on the last two lines? Substituting back into the derivative would give zero twice, because zero is the gradient, not the height.

Answer. (−1,3)(-1, 3) and (1,−1)(1, -1).

Check. Both horizontal coordinates give 3x2−3=03x^2-3=0, and both pairs satisfy the curve equation.

Change only the constant term from 11 to 33. The graph moves upwards by 22, while every gradient stays the same.

Worked example 2: a vertical shift

Question. Determine the stationary points of y=x3−3x+3y=x^3-3x+3.

Plan. Reuse the unchanged derivative, then calculate the new heights.

dydx=3x2−3constant shift differentiates to zero3x2−3=0⇒x=−1 or 1stationary coordinatesy(−1)=−1+3+3=5new heighty(1)=1−3+3=1new height\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-3 && \text{constant shift differentiates to zero}\\ 3x^2-3&=0\quad\Rightarrow\quad x=-1\text{ or }1 && \text{stationary coordinates}\\ y(-1)&=-1+3+3=5 && \text{new height}\\ y(1)&=1-3+3=1 && \text{new height} \end{aligned}

Why do the horizontal coordinates remain unchanged? Adding a constant changes heights, but not vertical differences or gradients.

Answer. (−1,5)(-1, 5) and (1,1)(1, 1).

Check. Each new point is 22 units above the corresponding point in Worked example 1.

A coordinate pair is the complete answer to a request for a point. Keep the derivative and curve labels separate.

Watch out

Wrong: “The stationary points are (−1,0)(-1, 0) and (1,0)(1, 0), because the derivative is zero.” Right: zero is the gradient; use the curve equation for each vertical coordinate.

Change the vertical shift once more.

Classifying with the second derivative

Differentiate the gradient function again to learn how the gradient itself is changing.

At a minimum, a smooth curve changes from falling to rising: its gradient changes from negative to positive. At a maximum, the change is from positive to negative. The second derivative measures how the first derivative changes, which explains its classification rule.

Second derivative

The second derivative is the derivative of the gradient function, written d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} or f′′(x)\mathrm{f}''(x). It is not the square of the first derivative. A local minimum is lower than nearby points on either side; a local maximum is higher than nearby points on either side.

For a polynomial at a stationary coordinate x=ax=a, a positive second derivative means the gradient is increasing through zero; a negative second derivative means it is decreasing through zero.

Second-derivative test

f′(a)=0Nature of the pointf′′(a)>0local minimumf′′(a)<0local maximumf′′(a)=0this test gives no conclusion\begin{array}{c|c} \mathrm{f}'(a)=0 & \text{Nature of the point}\\ \hline \mathrm{f}''(a)>0 & \text{local minimum}\\ \mathrm{f}''(a)<0 & \text{local maximum}\\ \mathrm{f}''(a)=0 & \text{this test gives no conclusion} \end{array}

Apply the sign test at a stationary point. A zero second derivative needs another method.
Not on the formulae sheet: learn it.

State the evaluated sign and what it means; a bare second derivative does not finish a classification.

Worked example 3: classifying both turning points

Question. The stationary points of y=x3−6x2+9x+2y=x^3-6x^2+9x+2 are (1,6)(1, 6) and (3,2)(3, 2). Determine their nature.

Plan. Find the second derivative, evaluate it at each stationary coordinate and interpret its sign.

dydx=3x2−12x+9first derivatived2ydx2=6x−12differentiate againf′′(1)=−6<0maximum at (1,6)f′′(3)=6>0minimum at (3,2)\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-12x+9 && \text{first derivative}\\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}&=6x-12 && \text{differentiate again}\\ \mathrm{f}''(1)&=-6<0 && \text{maximum at }(1,6)\\ \mathrm{f}''(3)&=6>0 && \text{minimum at }(3,2) \end{aligned}

Why evaluate at both coordinates? The second derivative's sign can change along the same curve.

Answer. (1,6)(1, 6) is a local maximum and (3,2)(3, 2) is a local minimum.

Check. The first derivative is 3(x−1)(x−3)3(x-1)(x-3), with signs positive, negative, positive in the three intervals. The curve rises, falls, then rises.

Change only the sign of the whole function. Reflecting it in the horizontal axis exchanges maxima and minima.

Worked example 4: a reflected curve

Question. Classify the stationary points (1,−6)(1, -6) and (3,−2)(3, -2) of y=−x3+6x2−9x−2y=-x^3+6x^2-9x-2.

Plan. Differentiate twice and use the sign at each given coordinate.

dydx=−3x2+12x−9first derivatived2ydx2=−6x+12second derivativef′′(1)=6>0minimum at (1,−6)f′′(3)=−6<0maximum at (3,−2)\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=-3x^2+12x-9 && \text{first derivative}\\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}&=-6x+12 && \text{second derivative}\\ \mathrm{f}''(1)&=6>0 && \text{minimum at }(1,-6)\\ \mathrm{f}''(3)&=-6<0 && \text{maximum at }(3,-2) \end{aligned}

Why do the signs reverse? Negating the function negates each derivative too, reversing rises and falls.

Answer. (1,−6)(1, -6) is a local minimum and (3,−2)(3, -2) is a local maximum.

Check. These are the reflected heights from Worked example 3, with the expected reversed classifications.

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