Why solve f prime of x equals 0 and not f of x equals 0?

Setting f of x equal to 0 finds where the curve crosses the x-axis, not where it is flat. Stationary points are about the gradient, so you must set the derivative f prime of x equal to 0. Solving f of x equals 0 instead is a common mistake.

Do you have to find the y-coordinate of a stationary point?

Yes. Solving f prime of x equals 0 only gives the x-coordinate. To state the stationary point fully you substitute that x back into the original function f of x to work out the matching y-coordinate.

Why solve f prime of x equals 0 and not f of x equals 0?

Setting f of x equal to 0 finds where the curve crosses the x-axis, not where it is flat. Stationary points are about the gradient, so you must set the derivative f prime of x equal to 0. Solving f of x equals 0 instead is a common mistake.

Do you have to find the y-coordinate of a stationary point?

Yes. Solving f prime of x equals 0 only gives the x-coordinate. To state the stationary point fully you substitute that x back into the original function f of x to work out the matching y-coordinate.

Year 12 Methods (Unit 3 & 4) Further differentiation and applications

Stationary points

Q: What is a stationary point?

A stationary point is a point on a curve where the gradient is zero, that is where the derivative f prime of x equals 0. The tangent there is horizontal. To find one, differentiate, set the derivative equal to zero and solve for x, then substitute back to get the y-coordinate.

Q: How do you classify a stationary point using the first derivative?

Test the sign of the derivative just to the left and just to the right of the stationary point. If it changes from positive to negative the point is a local maximum; from negative to positive it is a local minimum; if the sign does not change, the point is a stationary point of inflection.

Q: What is a stationary point of inflection?

It is a stationary point where the derivative is zero but does not change sign either side, so the curve is momentarily flat yet does not turn around. For example y equals x cubed has a stationary point of inflection at the origin: the gradient is positive on both sides of x equals 0.

Q: How many stationary points can a curve have?

As many as the number of real solutions of f prime of x equals 0. A quadratic has one, a cubic has up to two, and a quartic has up to three. For example y equals x to the fourth minus 2 x squared has three stationary points, at x equals minus 1, 0 and 1.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), a stationary point of a curve is a point where the gradient is zero, so \(f'(x)=0\) and the tangent is horizontal. You find one by solving \(f'(x)=0\), work out its \(y\)-coordinate, then classify its nature — a local maximum, a local minimum or a stationary point of inflection — using the first-derivative sign test.

A stationary point of a curve \(y=f(x)\) is a point where the gradient is zero: \(f'(x)=0\). At such a point the tangent is horizontal and the curve is momentarily flat.

Every stationary point is one of three types:

Local maximum — the curve rises then falls, so \(f'\) changes from \(+\) to \(-\).
Local minimum — the curve falls then rises, so \(f'\) changes from \(-\) to \(+\).
Stationary point of inflection — the curve is momentarily flat but does not turn around, so \(f'\) keeps the same sign either side.

The first-derivative sign test reads the sign of \(f'\) just to the left and just to the right of a stationary point to decide which type it is.

Key idea. Stationary point \(\Rightarrow f'(x)=0\). Classify by the sign of \(f'\) either side: \(+\) to \(-\) is a maximum, \(-\) to \(+\) is a minimum, no change is a stationary point of inflection.

\(f'\) turns \(+\) to \(-\) at the maximum \((1,5)\), and \(-\) to \(+\) at the minimum \((3,1)\)

\(y=x^{3}\): \(f'\) stays \(+\) either side of \(x=0\), so \((0,0)\) is a stationary point of inflection

A stationary point occurs where the derivative is zero:

\[f'(x)=0\]

f^{'} (x) = 0

The first-derivative sign test classifies each stationary point at \(x=a\):

\[f'(a^{-})>0,\ f'(a^{+})<0 \ \Rightarrow\ \text{local maximum}\]

\[f'(a^{-})<0,\ f'(a^{+})>0 \ \Rightarrow\ \text{local minimum}\]

\[f'(a^{-})\ \text{and}\ f'(a^{+})\ \text{same sign} \ \Rightarrow\ \text{stationary point of inflection}\]

f^{'} (a) = 0

Common derivatives used. \(\dfrac{d}{dx}\left(x^{n}\right)=nx^{n-1}\), \(\dfrac{d}{dx}e^{x}=e^{x}\), \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\), \(\dfrac{d}{dx}\sin x=\cos x\), \(\dfrac{d}{dx}\cos x=-\sin x\), together with the product and quotient rules.

How to find and classify stationary points

Differentiate. Find \(f'(x)\) using the standard rules (and the product or quotient rule where needed).
Solve \(f'(x)=0\). The solutions are the \(x\)-coordinates of the stationary points.
Find each \(y\)-coordinate. Substitute each \(x\) back into the original \(f(x)\) — not into \(f'(x)\).
Classify with a sign test. Check the sign of \(f'\) just left and just right of each stationary \(x\): \(+\) to \(-\) is a maximum, \(-\) to \(+\) is a minimum, no change is a stationary point of inflection.

Sign table. A neat way to record step 4 is a table with the stationary \(x\)-values as columns and the sign of \(f'\) in each interval, reading the shape (up-slope \(/\), flat, down-slope \(\backslash\)) straight off the signs.

Example 1 — Quadratic

Find and classify the stationary point of \(y=x^{2}-6x+5\).

Solution

Differentiate and solve \(y'=0\), then find \(y\) and sign-test.

\(y'\)	\(=\)	\(2x-6\)
\(2x-6\)	\(=\)	\(0\Rightarrow x=3\)
\(y(3)\)	\(=\)	\(9-18+5=-4\)

\(y'(2)=-2<0\), \(y'(4)=2>0\): \(-\) to \(+\), so a local minimum at \((3,-4)\).

local minimum (3, - 4)

Example 2 — Cubic, sign table

Find and classify the stationary points of \(y=x^{3}-6x^{2}+9x+1\).

Solution

\(y'\)	\(=\)	\(3x^{2}-12x+9=3(x-1)(x-3)\)
\(y'\)	\(=\)	\(0\Rightarrow x=1\ \text{or}\ 3\)

\(y(1)=5\), \(y(3)=1\). Sign test: \(y'(0)=9>0\), \(y'(2)=-3<0\), \(y'(4)=9>0\).

\((1,5)\) is a local maximum; \((3,1)\) is a local minimum.

(1, 5) max, (3, 1) min

Example 3 — Inflection

Classify the stationary point of \(y=x^{3}+1\).

Solution

\(y'\)	\(=\)	\(3x^{2}\)
\(3x^{2}\)	\(=\)	\(0\Rightarrow x=0\)
\(y(0)\)	\(=\)	\(1\)

\(y'(-1)=3>0\) and \(y'(1)=3>0\): no sign change, so \((0,1)\) is a stationary point of inflection.

inflection (0, 1)

Example 4 — With \(e^{x}\)

Find the stationary point of \(y=x\,e^{x}\) and state its nature.

Solution

Product rule, then factor out \(e^{x}\) (never \(0\)).

\(y'\)	\(=\)	\(e^{x}+x\,e^{x}=e^{x}(1+x)\)
\(e^{x}(1+x)\)	\(=\)	\(0\Rightarrow x=-1\)
\(y(-1)\)	\(=\)	\(-\dfrac{1}{e}\)

\(y'(-2)<0\), \(y'(0)=1>0\): \(-\) to \(+\), a local minimum at \(\left(-1,-\dfrac{1}{e}\right)\).

local minimum (- 1, - 1 / e)

Common pitfalls

Solve \(f'(x)=0\), not \(f(x)=0\). Stationary points are about the gradient, so set the derivative to zero. Solving \(f(x)=0\) finds \(x\)-intercepts, a different thing entirely.

Don't forget the \(y\)-coordinate. Solving \(f'(x)=0\) gives only the \(x\)-value; substitute it back into the original \(f(x)\) to state the point fully.

No sign change means inflection. If \(f'\) keeps the same sign either side, the point is a stationary point of inflection — not a maximum or minimum. Assuming it must be one or the other is a classic error.

Frequently asked questions

What is a stationary point?

A point where the gradient is zero, so \(f'(x)=0\) and the tangent is horizontal. Differentiate, solve \(f'(x)=0\), then substitute back for the \(y\)-coordinate.

How do you classify a stationary point using the first derivative?

Test the sign of \(f'\) just left and right of the point: \(+\) to \(-\) is a local maximum, \(-\) to \(+\) is a local minimum, and no change is a stationary point of inflection.

What is a stationary point of inflection?

A stationary point where \(f'=0\) but \(f'\) does not change sign, so the curve is flat for an instant yet does not turn. \(y=x^{3}\) has one at the origin.

Why solve \(f'(x)=0\) and not \(f(x)=0\)?

Because stationary points are where the curve is flat (zero gradient), which is \(f'(x)=0\). \(f(x)=0\) gives the \(x\)-intercepts instead.

Do you have to find the \(y\)-coordinate?

Yes. \(f'(x)=0\) gives only the \(x\)-coordinate; substitute it into \(f(x)\) to get the matching \(y\)-value and state the point.

How many stationary points can a curve have?

As many as the real solutions of \(f'(x)=0\): a quadratic has one, a cubic up to two, a quartic up to three. For instance \(y=x^{4}-2x^{2}\) has three.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

← Previous subtopic

Motion in a straight line

Next subtopic →

Types of stationary points

More in Further differentiation and applications