Curve sketching
Find stationary points and determine their nature using first and second derivative tests; identify concavity and points of inflection; determine increasing and decreasing intervals; sketch curves from derivative information.
Worked examples
Finding stationary points and their nature
Straightforward
Problem
For , find all stationary points and determine their nature.
1
Differentiate.
2
Solve .
3
Classify using the second derivative test.
At : , a local maximum. At : , a local minimum.
4
Find the -values.
Answer
Local maximum at and local minimum at .
Identifying points of inflection
Moderate
Problem
Find the point of inflection of and describe the concavity on either side.
1
Find .
2
Solve .
3
Check the sign change of either side of .
For , , so the curve is concave down. For , , so the curve is concave up. Concavity changes at , so there is a point of inflection.
4
Find the -value.
Answer
Point of inflection at .
Curve analysis with an exponential function
Challenging
Problem
For , find the stationary point and determine its nature.
1
Differentiate using the product rule with and .
2
Solve , noting for all .
3
Apply the second derivative test.
4
Find the -value.
Answer
Local maximum at .
Practise
Q1·Straightforward
A function has at and . What can you conclude about ?
Explanation
Since (stationary point) and (concave down), the second derivative test tells us is a **local maximum**.
Q2·Straightforward
Find the -coordinate of the stationary point of .
Explanation
.
Setting : .
The stationary point is at .
Setting : .
The stationary point is at .
Q3·Straightforward
For , find the value of at .
Explanation
Since , the second derivative test is inconclusive at — further investigation is needed.
Q4·Straightforward
A curve has on the interval . What does this mean?
Explanation
When , the gradient is increasing, and the curve is **concave up** (shaped like a cup ). This is independent of whether the curve is increasing or decreasing.
Q5·Moderate
For , find the -coordinate of the point of inflection.
Explanation
Setting : .
Check sign change: for , (concave down); for , (concave up). The concavity changes, so is a point of inflection.
Q6·Moderate
Find the -value at the local minimum of .
Explanation
Stationary points at and .
— local maximum at .
— local minimum at .
The local minimum value is .
Q7·Moderate
For , what is the -value at the local maximum?
Explanation
From differentiating: stationary points are at (local max) and (local min).
The local maximum value is .
The local maximum value is .
Q8·Moderate
The graph of is shown to cross the -axis from negative to positive at , and from positive to negative at . Which statement is correct about ?
Explanation
Using the first derivative test:
- At : changes from to , so changes from decreasing to increasing → **local minimum**.
- At : changes from to , so changes from increasing to decreasing → **local maximum**.
- At : changes from to , so changes from decreasing to increasing → **local minimum**.
- At : changes from to , so changes from increasing to decreasing → **local maximum**.
Q9·Moderate
Consider . On which interval is decreasing?
Explanation
Stationary points at and .
Sign of :
- : (increasing)
- : (decreasing)
- : (increasing)
So is decreasing on .
Q10·Moderate
For , find the -value of the point of inflection.
Explanation
Setting : .
For : (concave down). For : (concave up). Concavity changes, so there is a point of inflection at .
Q11·Challenging
The function has two stationary points. Find the sum of their -coordinates.
Explanation
Stationary points at and .
Sum .
(Alternatively, by Vieta's formulas for , sum of roots .)
Q12·Challenging
For , find the -coordinate of the stationary point. Give your answer as an integer.
Explanation
Using the product rule with and :
Since for all , setting gives , so .
, confirming a local minimum.
Since for all , setting gives , so .
, confirming a local minimum.
Q13·Challenging
For (defined for ), find the -coordinate of the stationary point.
Explanation
Setting : .
, confirming a local maximum at .
The maximum value is .
Open Math
Curve sketching
Calculus · MAV-12-06
Name:
Date:
Q1Straightforward
A function has at and . What can you conclude about ?
- A.Local minimum
- B.Local maximum
- C.Point of inflection
- D.Cannot be determined
Q2Straightforward
Find the -coordinate of the stationary point of .
Q3Straightforward
For , find the value of at .
Q4Straightforward
A curve has on the interval . What does this mean?
- A.The curve is decreasing on
- B.The curve is concave up on
- C.The curve is concave down on
- D.The curve has a stationary point on
Q5Moderate
For , find the -coordinate of the point of inflection.
Q6Moderate
Find the -value at the local minimum of .
Q7Moderate
For , what is the -value at the local maximum?
Q8Moderate
The graph of is shown to cross the -axis from negative to positive at , and from positive to negative at . Which statement is correct about ?
- A. has a local minimum at and a local maximum at
- B. has a local maximum at and a local minimum at
- C. has points of inflection at and
- D. is concave up at both and
Q9Moderate
Consider . On which interval is decreasing?
- A.
- B.
- C.
- D.
Q10Moderate
For , find the -value of the point of inflection.
Q11Challenging
The function has two stationary points. Find the sum of their -coordinates.
Q12Challenging
For , find the -coordinate of the stationary point. Give your answer as an integer.
Q13Challenging
For (defined for ), find the -coordinate of the stationary point.
Worked solutions and answers at openmath.au/year-12/advanced/applications-of-calculus/curve-sketching