What does f prime positive or negative tell you?

f prime greater than zero means f is increasing, sloping up. f prime less than zero means f is decreasing, sloping down.

What does f prime positive or negative tell you?

f prime greater than zero means f is increasing, sloping up. f prime less than zero means f is decreasing, sloping down.

Year 11 Methods (Unit 1 & 2) Differentiation Of Polynomials

Graphs Of The Derivative Function

Q: How do you tell if a graph is f or its derivative?

The derivative crosses the x-axis where f has a stationary point, and is positive where f rises. It is also one degree lower than f.

Q: What does the x-intercept of f prime of x mean?

It marks a stationary point of f: the x-value where f has a horizontal tangent, a maximum, minimum or stationary point of inflection.

Q: How do you tell a maximum from a minimum using f prime?

If f prime changes from positive to negative, f has a maximum there. From negative to positive, a minimum.

Q: Why is the derivative graph one degree lower?

The power rule lowers each power by one, so a quadratic gives a linear derivative and a cubic gives a quadratic derivative.

20 practice questions 1 video lesson Theory + worked examples

Understand graphs of the derivative function for Queensland Year 11 Mathematical Methods (QCAA). The derivative graph records the gradient of the original curve: positive where it rises, negative where it falls, and zero at stationary points.

You will learn to sketch the derivative graph from a curve, match a function to its derivative, and read where a function increases or decreases — building a strong visual sense of gradient.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 1 video(s)

Graphs Of The Derivative Function - Video - Graphs of derivative functions Watch

Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), the graph of the derivative \(y=f'(x)\) records the gradient of \(y=f(x)\) at every point. Where \(f\) is increasing, \(f'>0\); where it is decreasing, \(f'<0\); and at a stationary point, \(f'=0\). This page shows how to read one graph from the other and how to decide which graph is \(f'\).

The gradient function \(f'(x)\) turns the slope of \(y=f(x)\) into a height. So the graph of \(y=f'(x)\) is positive exactly where \(f\) rises, negative where \(f\) falls, and zero where \(f\) has a horizontal tangent.

The \(x\)-intercepts of \(y=f'(x)\) (its zeros) sit directly below the stationary points of \(y=f(x)\). A maximum of \(f\) shows as \(f'\) changing from \(+\) to \(-\); a minimum as \(f'\) changing from \(-\) to \(+\).

Differentiating lowers the degree by one: a quadratic \(f\) has a linear \(f'\); a cubic \(f\) has a quadratic \(f'\). That shape clue helps you match a curve to its derivative.

Slope of \(f\) becomes height of \(f'\). Read \(f\) left to right: uphill means \(f'\) above the axis, downhill means below, flat means on the axis.

\(y=x^2-4x\) (navy) is stationary at \(x=2\), exactly where \(f'=2x-4\) (gold) is zero.

The turning points of \(y=x^3-3x\) at \(x=\pm1\) are the zeros of \(f'=3x^2-3\).

The sign of the derivative reads the shape of the curve:

\[f'(x)>0\ ext{increasing},\quad f'(x)<0\ ext{decreasing},\quad f'(x)=0\ ext{stationary}\]

f^{'} (x) > 0

The degree drops by one when you differentiate:

\[\text{quadratic }f\ \Rightarrow\ \text{linear }f',\qquad ext{cubic }f\ \Rightarrow\ ext{quadratic }f'\]

\deg f^{'} = \deg f - 1

Zeros of \(f'\) sit under stationary points of \(f\). Sign change \(+\!\to\!-\) marks a maximum; \(-\!\to\!+\) marks a minimum.

Reading \(y=f'(x)\) from \(y=f(x)\) (and matching)

Mark the flat spots: the stationary points of \(f\) become the \(x\)-intercepts of \(f'\).
Sign the intervals: where \(f\) rises put \(f'\) above the axis; where \(f\) falls put \(f'\) below.
Check the degree: \(f'\) is one degree lower than \(f\) — use this to reject wrong graphs.
Confirm a point: pick one \(x\) and check the sign (and, if easy, the value) of \(f'\) matches.

Example 1 — Where is \(f\) increasing or decreasing?

For \(y=x^2-4x\), find where the curve is increasing, decreasing and stationary, and give \(f'(x)\).

Solution

Differentiate:

\(f'(x)\)

\(=\)

\(2x-4\)

Stationary point — solve \(f'(x)=0\):

\(2x-4\)	\(=\)	\(0\)
\(x\)	\(=\)	\(2\)

Sign of \(f'\) either side:

\(f'(0)\)	\(=\)	\(2(0)-4=-4<0\)
\(f'(4)\)	\(=\)	\(2(4)-4=4>0\)

So \(f'<0\) for \(x<2\) (decreasing) and \(f'>0\) for \(x>2\) (increasing).

Decreasing for \(x<2\), increasing for \(x>2\), stationary (minimum) at \(x=2\).

f^{'} (x) = 2 x - 4

Example 2 — Zero of \(f'\) locates the stationary point

The gradient function of a curve is \(f'(x)=2x-6\). Where is the stationary point, and is \(f\) increasing at \(x=5\)?

Solution

Stationary point — the \(x\)-intercept of \(f'\):

\(2x-6\)	\(=\)	\(0\)
\(x\)	\(=\)	\(3\)

Sign of \(f'\) at \(x=5\):

\(f'(5)\)	\(=\)	\(2(5)-6\)
	\(=\)	\(4\)
	\(=\)	\(>0\)

\(f'(5)>0\), so \(f\) is increasing at \(x=5\).

Stationary point at \(x=3\); \(f\) is increasing at \(x=5\) (since \(f'(5)=4>0\)).

x = 3

Example 3 — Which graph is \(f'\)?

A cubic is \(f(x)=x^3-3x\). Describe the graph of \(y=f'(x)\) and where it meets the \(x\)-axis.

Solution

Differentiate (degree drops from 3 to 2):

\(f'(x)\)

\(=\)

\(3x^2-3\)

Zeros of \(f'\) — the stationary points of \(f\):

\(3x^2-3\)	\(=\)	\(0\)
\(x^2\)	\(=\)	\(1\)
\(x\)	\(=\)	\(\pm 1\)

So \(y=f'(x)\) is an upward parabola cutting the \(x\)-axis at \(x=-1\) and \(x=1\); \(f'<0\) between them (where \(f\) falls) and \(f'>0\) outside.

\(y=f'(x)=3x^2-3\): an upward parabola with \(x\)-intercepts at \(x=\pm1\).

f^{'} (x) = 3 x^{2} - 3

Example 4 — Reasoning from \(f'\) back to \(f\)

The gradient function of a curve is the line \(f'(x)=2x-2\). Describe the shape of \(y=f(x)\).

Solution

Find the stationary \(x\)-value — where \(f'=0\):

\(2x-2\)	\(=\)	\(0\)
\(x\)	\(=\)	\(1\)

Sign of \(f'\) either side of \(x=1\):

\(f'(0)\)	\(=\)	\(-2<0\)
\(f'(2)\)	\(=\)	\(2>0\)

\(f\) falls for \(x<1\) and rises for \(x>1\), so the turning point at \(x=1\) is a minimum. Since \(f'\) is linear, \(f\) is a quadratic (an upward parabola).

\(y=f(x)\) is an upward parabola with a minimum at \(x=1\).

x = 1

Common pitfalls

Reading height instead of slope. \(f'(x)\) is the slope of \(f\), not its height. A high point on \(f\) does not mean a high point on \(f'\).

Mixing up the two graphs. Where \(f\) is decreasing, \(f'\) is below the axis (negative), even though \(f\) itself may still be positive.

Ignoring the degree clue. A cubic \(f\) must have a quadratic \(f'\); a graph of \(f'\) with the wrong shape can be ruled out at once.

Forgetting the sign change tells max vs min. \(f'\) going \(+\!\to\!-\) is a maximum of \(f\); \(-\!\to\!+\) is a minimum.

Frequently asked questions

How do you tell if a graph is f or its derivative?

The derivative \(f'\) crosses the \(x\)-axis where \(f\) has a stationary point, and is positive where \(f\) rises. It is also one degree lower than \(f\).

What does the x-intercept of \(f'(x)\) mean?

It marks a stationary point of \(f\): the \(x\)-value where \(f\) has a horizontal tangent (a maximum, minimum or stationary point of inflection).

What does f' positive or negative tell you?

\(f'>0\) means \(f\) is increasing (sloping up); \(f'<0\) means \(f\) is decreasing (sloping down).

How do you tell a maximum from a minimum using f'?

If \(f'\) changes from positive to negative, \(f\) has a maximum there; from negative to positive, a minimum.

Why is the derivative graph one degree lower?

The power rule lowers each power by one, so a quadratic gives a linear derivative and a cubic gives a quadratic derivative.

← Previous subtopic

Rules For Differentiation

Next subtopic →

Limits And Continuity

Graphs Of The Derivative Function

📖 Theory

Reading \(y=f'(x)\) from \(y=f(x)\) (and matching)

Common pitfalls

Frequently asked questions

Theory