P2 7.1 - Stationary points and curve shape | Edexcel A-Level International Mathematics

Use the sign of a derivative to decide where a curve rises or falls. Find stationary points, distinguish maxima and minima from flat points that do not turn, and combine this information with intercepts to sketch a curve.

Increasing and decreasing intervals

Must a horizontal tangent mark a turn? The curve $y=x^3$ passes through $(-1,-1)$ , $(0,0)$ and $(1,1)$ : it keeps rising even though its gradient is zero at the origin. A gradient tells you about change, not height.

Move from left to right along $y=x^2$ . Before the origin its tangent slopes down; afterwards its tangent slopes up. Its derivative $2x$ has exactly those signs.

Increasing and decreasing

A function is increasing on an interval if larger inputs give larger outputs, and decreasing if larger inputs give smaller outputs. For a differentiable function, $f'(x)>0$ throughout an interval establishes that it is increasing there; $f'(x)<0$ establishes that it is decreasing there.

Find the inputs where the derivative is zero or undefined, then determine its sign on each remaining interval. A factorised derivative makes this easier. State intervals in terms of $x$ , rather than saying that the function is increasing “when $y>0$ ”.

Worked example 1: read a derivative's sign

Question. Find the intervals on which $f(x)=x^2-4$ is increasing or decreasing.

Plan. Differentiate and solve the two sign inequalities.

\begin{aligned} f'(x)&=2x\\ 2x<0&\iff x<0\\ 2x>0&\iff x>0 \end{aligned}

Answer. Decreasing for $x<0$ ; increasing for $x>0$ .

Check. $f(-2)=0>f(-1)=-3$ , while $f(1)=-3<f(2)=0$ . These sample changes agree with the derivative analysis.

Change only the sign of the squared term. The graph's rise and fall reverse, although its height at the origin stays the same.

Worked example 2: reverse the leading sign

Question. Find the increasing and decreasing intervals of $g(x)=-x^2-4$ .

Plan. Differentiate, remembering that division by a negative number reverses an inequality.

\begin{aligned} g'(x)&=-2x\\ -2x>0&\iff x<0\\ -2x<0&\iff x>0 \end{aligned}

Answer. Increasing for $x<0$ ; decreasing for $x>0$ .

Check. $g(-2)=-8<g(-1)=-5$ . The function is increasing here even though both values are negative.

A zero derivative at one isolated input does not mean the function is constant on an interval. For $x^3$ , $3x^2$ is positive on both sides of zero. Also respect the original domain: an excluded input is not a point on the curve.

Watch out

$f(x)>0$ describes a position above the axis. $f'(x)>0$ describes an upward slope. Neither statement implies the other.

Use a new coefficient while keeping the same sign decision.

Stationary points

A stationary point has a horizontal tangent, so its gradient is zero; finding it requires both an input and an output.

Stationary point

A stationary point of $y=f(x)$ is a point on the curve at which $f'(x)=0$ . Its coordinates are $(a,f(a))$ , where $a$ solves $f'(a)=0$ .

Differentiate first, solve the resulting equation, then substitute each valid root into the original function. Substitution into the derivative only confirms that the gradient is zero; it cannot give the height.

Worked example 3: two stationary points

Question. Find the stationary points of $f(x)=x^3-3x^2+2$ .

Plan. Factorise the derivative, solve for its zeros, then find the corresponding heights.

\begin{aligned} f'(x)&=3x^2-6x=3x(x-2)\\ f'(x)=0&\implies x=0\text{ or }x=2\\ f(0)&=2\\ f(2)&=8-12+2=-2 \end{aligned}

Answer. $(0,2)$ and $(2,-2)$ .

Check. The derivative is zero at both inputs: $3(0)(-2)=0$ and $3(2)(0)=0$ . Direct substitution into the original cubic gives the two stated heights.

Change only the linear coefficient, from zero to $3$ . The derivative now has a repeated root, which gives one stationary point rather than two distinct points.

Worked example 4: a repeated derivative root

Question. Find the stationary points of $g(x)=x^3-3x^2+3x+2$ .

Plan. Complete the derivative's square, then substitute its root into $g$ .

\begin{aligned} g'(x)&=3x^2-6x+3=3(x-1)^2\\ g'(x)=0&\implies x=1\\ g(1)&=1-3+3+2=3 \end{aligned}

Answer. The only stationary point is $(1,3)$ .

Check. Rewriting the original function gives $g(x)=(x-1)^3+3$ . This is a translated cubic with its horizontal tangent at $(1,3)$ .

The equation $g'(x)=0$ locates the point; it has not established a maximum or minimum. That requires the behaviour around it.

Watch out

Do not divide $3x(x-2)=0$ by $x$ : that would lose the solution $x=0$ . Set each factor equal to zero.

Try another derivative with a repeated root.

Classifying maxima and minima

A curve turns at a stationary point only if its gradient changes sign there.

A local maximum is at least as high as nearby points; a local minimum is at least as low as nearby points. “Local” makes no claim about distant points or the ends of a restricted interval.

Sign of $f'$ just before and after	Behaviour at the stationary point
Positive, then negative	Local maximum
Negative, then positive	Local minimum
Same sign on both sides	Neither a maximum nor a minimum

There is also a shortcut using the second derivative. If $f''(a)>0$ , the gradient is increasing through zero: negative before, positive after. If $f''(a)<0$ , it decreases through zero.

Second derivative test

At a stationary point $x=a$ , for the smooth functions considered here,

f''(a)>0\implies\text{local minimum},\qquad f''(a)<0\implies\text{local maximum}.

If $f''(a)=0$ , this test is inconclusive: examine the sign of $f'$ on either side instead.

Not in the formula booklet: learn it.

Start with a cubic whose two stationary points give nonzero second derivatives.

Worked example 5: classify using the second derivative

Question. Classify the stationary points of $f(x)=x^3-3x+1$ .

Plan. Find the stationary inputs, then evaluate the second derivative at each.

\begin{aligned} f'(x)&=3x^2-3=3(x-1)(x+1)\\ f'(x)=0&\implies x=-1,\ 1\\ f''(x)&=6x\\ f''(-1)&=-6<0,\qquad f(-1)=3\\ f''(1)&=6>0,\qquad f(1)=-1 \end{aligned}

Answer. A local maximum at $(-1,3)$ , and a local minimum at $(1,-1)$ .

Check. The signs of $3(x-1)(x+1)$ are positive, negative, positive on the three intervals separated by $-1$ and $1$ . They confirm both classifications independently.

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P2 7.1 - Stationary points and curve shape

Increasing and decreasing intervals

Stationary points

Classifying maxima and minima

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