Two Tests, Different Information
A critical number is a point in the domain where the derivative is zero or does not exist. Once a critical number is found, the next question is what the function does nearby: does it have a local maximum, a local minimum, or neither?
Two common tools answer this question in different ways. The first derivative test examines whether the function is increasing or decreasing on each side of the critical number. The second derivative test uses the curvature indicated by the second derivative at a stationary point. The first test describes nearby changes in direction; the second can often classify a point with less sign-chart work, but only under its conditions.
What Each Test Requires
For an interior point \(c\), the first derivative test applies when \(f\) is continuous at \(c\) and differentiable on intervals immediately to its left and right. The derivative need not exist at \(c\) itself. What matters is whether the derivative has a consistent sign on each side, close enough to \(c\).
The phrase “just to the left and right” means on some intervals immediately adjacent to \(c\), not necessarily on the whole domain. A sign change from positive to negative corresponds to increasing and then decreasing; a change from negative to positive corresponds to decreasing and then increasing.
The second derivative test is narrower. It requires a stationary point, \(f'(c)=0\), and a second derivative at \(c\). If that second derivative is nonzero, its sign classifies the point.
Here, the conditions for the test are local: \(f'\) must be defined in a neighborhood of \(c\), so that \(f''(c)\), the derivative of \(f'\) at \(c\), is defined. In practice, differentiating \(f'\) and evaluating at \(c\) is often the shortest route when the test applies.
Why the Second Derivative Test Works
The second derivative measures the instantaneous rate of change of the first derivative. At a stationary point, \(f'(c)=0\). If \(f''(c)>0\), then near \(c\), the derivative changes approximately like a positive multiple of \(x-c\): it is negative to the left and positive to the right. That is exactly the decrease-then-increase pattern associated with a local minimum.
This reasoning can be made precise using the definition of \(f''(c)\). Because \(f'(c)=0\),
If \(f''(c)>0\), then the quotient \(\frac{f'(x)}{x-c}\) is positive for all \(x\ne c\) sufficiently close to \(c\). Thus \(f'(x)<0\) to the left of \(c\), and \(f'(x)>0\) to the right. The Mean Value Theorem then turns those derivative signs into comparisons of function values.
For \(x\) close to \(c\), \(f'(x)/(x-c)>0\) when \(f''(c)>0\). Hence \(f'(x)<0\) left of \(c\), and \(f'(x)>0\) right of \(c\).
For \(x<c\) sufficiently close to \(c\), the Mean Value Theorem gives \(f(c)-f(x)=f'(\xi)(c-x)\) for some \(\xi\) between \(x\) and \(c\). Since \(f'(\xi)<0\) and \(c-x>0\), \(f(c)-f(x)<0\), so \(f(x)>f(c)\).
For \(x>c\) sufficiently close to \(c\), the theorem gives \(f(x)-f(c)=f'(\xi)(x-c)\) for some \(\xi\) between \(c\) and \(x\). Both factors are positive, so \(f(x)>f(c)\).
Values on both sides of \(c\) are therefore greater than \(f(c)\), proving a local minimum. If \(f''(c)<0\), the quotient is negative near \(c\); \(f'\) is positive on the left and negative on the right, and the same Mean Value Theorem argument shows that nearby function values are less than \(f(c)\). This proves the local maximum case.
Worked Example: Both Tests Agree
Worked Example: Classifying Two Critical Numbers of a Cubic
Let \(f(x)=x^3-3x\). Find and classify its critical numbers using both tests.
Differentiate and factor:
Setting \(f'(x)=0\) gives \(x=1\) and \(x=-1\). The derivative is defined everywhere, so these are the critical numbers. For the first derivative test, the factors show \(f'(x)>0\) when \(x<-1\), \(f'(x)<0\) when \(-1<x<1\), and \(f'(x)>0\) when \(x>1\). Therefore, \(f'\) changes from positive to negative at \(-1\), and from negative to positive at \(1\). The first test gives a local maximum at \(-1\) and a local minimum at \(1\).
For the second derivative test,
Since \(f'(-1)=f'(1)=0\), the test applies at both points. It gives a local maximum at \(-1\) and a local minimum at \(1\), matching the first test. The function values are \(f(-1)=(-1)^3-3(-1)=-1+3=2\) and \(f(1)=1^3-3(1)=1-3=-2\). The extrema are \((-1,2)\) and \((1,-2)\). As a check, differentiating \(x^3-3x\) gives \(3x^2-3\), and differentiating again gives \(6x\), as used above.
When the First Test Gives More
The second derivative test can be inconclusive when \(f''(c)=0\). This does not mean that the point is not an extremum; it means only that this particular test does not decide. Checking the signs of \(f'\) on either side can still settle the question.
Worked Example: The Second Derivative Test Is Inconclusive
Let \(f(x)=x^4\). Classify the critical number at \(x=0\).
Differentiate:
The derivative is zero at \(x=0\), and \(f'(x)<0\) for \(x<0\) while \(f'(x)>0\) for \(x>0\). Thus the first derivative test gives a local minimum at \(0\). The function value is \(f(0)=0^4=0\).
Now differentiate again:
The second derivative test is inconclusive. This is not a contradiction: that test supplies no classification when \(f''(0)=0\), while the first derivative test does. Direct substitution also checks the classification: for every \(x\ne0\), \(f(x)-f(0)=x^4-0=x^4>0\). Therefore \(f(0)\) is less than nearby function values. Differentiating \(x^4\) gives \(4x^3\), and differentiating \(4x^3\) gives \(12x^2\), verifying both derivatives.
When the Second Derivative Test Is Unavailable
The second derivative test cannot classify a critical number where \(f'\) does not exist, because it requires \(f'(c)=0\) and \(f''(c)\) to exist. The first derivative test may still apply if \(f\) is continuous at the point and the derivative has definite signs on the two sides.
Worked Example: A Critical Point Where the Second Test Cannot Be Used
Let \(f(x)=|x|\). For \(x<0\), \(f(x)=-x\), so \(f'(x)=-1<0\). For \(x>0\), \(f(x)=x\), so \(f'(x)=1>0\). The derivative is undefined at \(0\), which is in the domain, so \(0\) is a critical number.
The derivative changes from negative to positive there. By the first derivative test, \(f\) has a local minimum at \(0\), with \(f(0)=0\). Indeed, for every \(x\ne0\), \(|x|>0=f(0)\), which directly checks the conclusion. The second derivative test is unavailable because \(f'(0)\) does not exist. On each side, the derivative calculation checks out: differentiating \(-x\) gives \(-1\), and differentiating \(x\) gives \(1\).
Choosing a Method and Avoiding Overclaims
When \(f'(c)=0\) and \(f''(c)\ne0\), the second derivative test is often an efficient way to classify \(c\). Use the first derivative test when you need the behavior on both sides, when \(f''(c)=0\), or when the derivative is undefined at \(c\) but exists on either side. A derivative sign chart may also reveal intervals of increase and decrease, information that the second derivative test at one point does not provide.
A common mistake is to treat an inconclusive result as a classification. For \(f(x)=x^4\), the incorrect reasoning is:
The implication is invalid. The second derivative test says nothing decisive when \(f''(0)=0\). The correction is to use another valid method: here \(f'(x)=4x^3\) is negative to the left of zero and positive to the right, so \(f\) has a local minimum.
Both tests give local conclusions. To claim an absolute extremum on a closed interval, for example, you must compare function values at all relevant critical numbers and endpoints. Neither a positive \(f''(c)\) nor a negative-to-positive sign change by itself compares \(f(c)\) with values far away.
Check Your Understanding
For each question, identify which test applies and state what conclusion it supports.
- If \(f'(c)=0\) and \(f''(c)<0\), what does the second derivative test conclude?
- If \(f'\) changes from positive to negative across \(c\), what does the first derivative test conclude?
- Why does \(f''(c)=0\) fail to classify \(c\) using the second derivative test?
- Can the second derivative test be used at a critical number where \(f'(c)\) does not exist? Explain.
- Does a local minimum established by either test necessarily have the least function value on the entire domain?
