9D, 9E - Stationary and turning points

A horizontal tangent can mark the top of a hill, the bottom of a valley, or a place where a curve briefly flattens and carries on. You will find stationary points, justify their nature and distinguish a turning point from a stationary point without a turn.

Finding stationary points

If a curve has gradient zero, must it turn back? First we need to locate the points where this question arises.

For $y=x^3-3x+2$ , the derivative is $3x^2-3$ . It is zero at $x=-1$ and $x=1$ . These are horizontal-tangent positions, but a position on the curve needs both coordinates.

Stationary point

A stationary point is a point on a curve where the derivative exists and is zero. For $y=f(x)$ , solve $f'(x)=0$ , then substitute each solution into $f(x)$ to find its $y$ -coordinate.

Do not solve $f(x)=0$ : that finds crossings of the horizontal axis. Also check that each candidate belongs to the domain of the original function. A value excluded by a denominator cannot become a point on the curve.

Worked example 1: finding both coordinates

Question. Find the stationary points of $y=x^3-3x+2$ .

Plan. Differentiate, solve the zero-gradient equation and recover the heights from the original equation.

\begin{aligned} \frac{dy}{dx}&=3x^2-3 &&\text{differentiate each term}\\ 0&=3(x-1)(x+1) &&\text{set the gradient to zero}\\ x&=-1\text{ or }1 &&\text{solve the two factors} \end{aligned}

At $x=-1$ , $y=-1+3+2=4$ . At $x=1$ , $y=1-3+2=0$ .

Answer. $(-1,4)$ and $(1,0)$ .

Check. Both values give $3x^2-3=0$ ; substituting both coordinate pairs into the original curve also works. The point $(-1,4)$ shows why stationary points need not lie on the horizontal axis.

Now change only the coefficient of the linear term. The zero-gradient positions move, so the heights must be calculated again.

Worked example 2: changing the linear term

Question. Find the stationary points of $y=x^3-12x+2$ .

Plan. Use the same zero-gradient method with the new derivative.

\frac{dy}{dx}=3x^2-12=3(x-2)(x+2)=0 \quad\Longrightarrow\quad x=-2\text{ or }2.

Substitution into the curve gives

y(-2)=-8+24+2=18,\qquad y(2)=8-24+2=-14.

Answer. $(-2,18)$ and $(2,-14)$ .

Check. The derivative at both positions is $3(4)-12=0$ . Also $y-2=x^3-12x$ is odd, so the two heights should be equally far above and below $2$ : they are $18$ and $-14$ .

Use the derivative for positions and the original function for heights.

Classifying stationary points

A turning point is identified by what the gradient does on the two sides of it, rather than by its height above the axis.

At a local maximum, nearby curve values are no greater: the curve changes from increasing to decreasing. At a local minimum, nearby values are no smaller: the curve changes from decreasing to increasing. “Local” compares nearby points; it does not promise the greatest or least value on an entire domain.

Tests for a stationary point

At a stationary point with $x=a$ :

Gradient just before $a$	Gradient just after $a$	Nature
positive	negative	local maximum
negative	positive	local minimum
same sign	same sign	no turn

The second derivative, $f''(x)$ , is the derivative of $f'(x)$ . For the smooth functions here, $f''(a)>0$ establishes a local minimum and $f''(a)<0$ establishes a local maximum. If $f''(a)=0$ , this test is inconclusive; examine the first-derivative signs.

The second derivative measures how the gradient is changing. A positive value at a stationary point means the gradient is increasing through zero; a negative value means it is decreasing through zero.

Worked example 3: a maximum and a minimum

Question. Classify the stationary points $(-1,4)$ and $(1,0)$ of $f(x)=x^3-3x+2$ .

Plan. Evaluate the second derivative at each stationary position.

f'(x)=3x^2-3,\qquad f''(x)=6x.

At $x=-1$ , $f''(-1)=-6<0$ , so $(-1,4)$ is a local maximum. At $x=1$ , $f''(1)=6>0$ , so $(1,0)$ is a local minimum.

Answer. Local maximum at $(-1,4)$ ; local minimum at $(1,0)$ .

Check. The factorisation $f'(x)=3(x-1)(x+1)$ gives signs $+, -, +$ on the intervals separated by $-1$ and $1$ . These are precisely the required changes.

Reflect the curve in the horizontal axis by multiplying its function by $-1$ . The stationary positions stay fixed, but the turns reverse.

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9D, 9E - Stationary and turning points

Finding stationary points

Classifying stationary points

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