How can the second derivative describe concavity?
Quick answer
For a function that is twice differentiable on an interval, a positive second derivative throughout means the graph is concave up there; a negative second derivative throughout means concave down. [S1] Find the second derivative and check its sign on each interval. A test point can help determine the sign on an interval, but a single point alone does not establish concavity across that interval.
The question, the details, the next step.
How do I use the second derivative to tell whether a graph is concave up or down?
Think of the second derivative as describing how the graph bends over an interval. Where f″(x) is positive throughout an interval, the function is concave up there; where f″(x) is negative throughout, it is concave down. [S1]
A practical check is:
- Find f″(x).
- Identify points in the domain where f″(x)=0 or where f″ is undefined. These may divide the domain into intervals to check.
- Determine the sign of f″ on each interval.
- State concavity separately for each interval.
The phrase “throughout an interval” matters: knowing the sign at just one point does not, by itself, establish the sign everywhere on that interval. [S1]
Can you work through a polynomial example and explain what the signs mean?
Take f(x)=x³. Its first derivative is f′(x)=3x², and its second derivative is f″(x)=6x. [S1]
The second derivative is zero at x=0, so check the intervals on either side:
- If x is in (−1,0), then 6x<0. Thus f is concave down on (−1,0).
- If x is in (0,1), then 6x>0. Thus f is concave up on (0,1). [S1]
In graph language, the curve bends downward on the first interval and upward on the second. The calculation at x=0 gives f″(0)=0, but that single value does not tell you the concavity on either interval. It is the sign across each interval that gives the conclusion.
Why isn’t testing one point enough? How should I choose a test point?
A test point can help you determine the sign on an interval when you have reason to know the sign stays the same throughout that interval. In this example, f″(x)=6x is negative throughout (−1,0) and positive throughout (0,1). You could choose any convenient point from each interval: for instance, x=−0.5 gives f″(−0.5)=−3, while x=0.5 gives f″(0.5)=3.
Those calculations illustrate the sign on the intervals shown. Don’t use a test point from one interval to make a claim about another. Also consider the function’s domain and places where the second derivative is undefined when deciding which intervals to check. The sign rule applies on intervals where the function is twice differentiable. [S1]
What can I write as a clear final answer for a homework problem?
Write the second derivative, identify the intervals you checked, and state the concavity on each. For the worked example, a concise answer is:
“f(x)=x³, so f″(x)=6x. The second derivative is negative on (−1,0) and positive on (0,1). Therefore, f is concave down on (−1,0) and concave up on (0,1). At x=0, f″(0)=0; the interval sign checks, rather than that value alone, establish the concavity on either side.”
This method establishes the concavity conclusions on the intervals checked. State those intervals explicitly rather than treating a single value as a conclusion about the graph’s bending.
A practical checklist
Compute f″(x) and simplify it.
Identify domain points where f″ is zero or undefined.
Use those points to split the domain into intervals.
Check the sign of f″ on each interval, using a test point when helpful.
State concave up or concave down for each interval; do not infer interval behavior from one isolated value.
A few more useful answers
If f″(a)=0, is the graph automatically changing concavity at a?
No. A zero at one point does not establish concavity on either side or prove that the concavity changes. Check the second-derivative sign on intervals around the point.
Can the second derivative rule be used wherever f″ exists at a point?
The interval rule concerns a function that is twice differentiable on the interval being considered. A value at one point is not enough to establish the sign throughout an interval. [S1]
What if the second derivative is undefined at a point?
Check whether that point is in the function’s domain, then split the domain into intervals around it as appropriate. Evaluate concavity on intervals where the function is twice differentiable.
Sources
References used for this guide.
- [S1] OpenStax Calculus Volume 1, §4.5: Derivatives and the Shape of a Graph openstax.org · Checked