Sketching Curves Using Stationary Points | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Put coordinates and directions together

A sketch shows the important features, not every calculated point. Find the intercepts, stationary points and direction between them. Keep any breaks in the domain.

Worked example

Sketch y = x³ − 3x.

On the $x$ -axis, $y=0$ : $x(x^2-3)=0$ gives $x=0,\pm\sqrt3$ . The $y$ -intercept is also zero.
$y'=3(x-1)(x+1)$ : maximum $(-1,2)$ , minimum $(1,-2)$ .
The derivative signs are positive, negative, positive, so connect the points by rising, falling, then rising again.

For large $|x|$ , the cubic term dominates the linear term. The left end goes down and the right end goes up.

The two plots share the x scale. Their vertical scales differ: curve height and gradient are different quantities.

If the graph given is the derivative

Read the vertical label first. A zero of $f'$ gives a stationary input of $f$ , not necessarily an intercept of $f$ . Where the derivative graph is below its axis, the original function decreases.

The derivative does not give the original curve height. Curves differing only by an added constant have the same derivative. Use a supplied point or the original equation to locate the sketch vertically.

Your turnPractice

A derivative graph crosses its axis at $x=-2$ and $x=2$ . It is positive outside these values and negative between them. The original curve passes through $(-2,16)$ and $(2,-16)$ . Classify these points and describe the curve between them.

Show worked answer

Positive to negative gives a maximum at $(-2,16)$ ; negative to positive gives a minimum at $(2,-16)$ . The curve decreases between them. Its heights come from the supplied points, not from the derivative graph.

Draw the sketch yourself

Your turnPractice

Sketch $y=x^3-12x$ , labelling its intercepts and stationary points. Use derivative signs to check the direction of every section.

Show worked answer

$y'=3(x-2)(x+2)$ . The maximum is $(-2,16)$ and the minimum is $(2,-16)$ . Intercepts: $x=0,\pm2\sqrt3$ . Draw rise–fall–rise, with the left end down and right end up.

Put coordinates and directions together

A sketch shows the important features, not every calculated point. Find the intercepts, stationary points and direction between them. Keep any breaks in the domain.

Worked example

Sketch y = x³ − 3x.

On the $x$ -axis, $y=0$ : $x(x^2-3)=0$ gives $x=0,\pm\sqrt3$ . The $y$ -intercept is also zero.
$y'=3(x-1)(x+1)$ : maximum $(-1,2)$ , minimum $(1,-2)$ .
The derivative signs are positive, negative, positive, so connect the points by rising, falling, then rising again.

For large $|x|$ , the cubic term dominates the linear term. The left end goes down and the right end goes up.

The two plots share the x scale. Their vertical scales differ: curve height and gradient are different quantities.

If the graph given is the derivative

Read the vertical label first. A zero of $f'$ gives a stationary input of $f$ , not necessarily an intercept of $f$ . Where the derivative graph is below its axis, the original function decreases.

Your turnPractice

Show worked answer

Draw the sketch yourself

Your turnPractice

Sketch $y=x^3-12x$ , labelling its intercepts and stationary points. Use derivative signs to check the direction of every section.

Show worked answer

$y'=3(x-2)(x+2)$ . The maximum is $(-2,16)$ and the minimum is $(2,-16)$ . Intercepts: $x=0,\pm2\sqrt3$ . Draw rise–fall–rise, with the left end down and right end up.