From Curvature to a Local Extremum
A stationary point is a point where the first derivative is zero. The second derivative test uses the value of the second derivative there to determine whether the function has a local minimum or maximum. Its logic is more precise than the informal idea that a graph “curves upward” or “curves downward”: the sign of \(f''(c)\) tells us how \(f'\) behaves close to the stationary point, and the sign of \(f'\) tells us how \(f\) behaves.
The key step is to use the definition of \(f''(c)\). If \(f''(c)>0\), then close to \(c\), the slope of \(f'\) is positive. Since \(f'(c)=0\), this forces \(f'\) to be negative just to the left of \(c\) and positive just to the right. The function therefore decreases before \(c\) and increases after it, giving a local minimum. A negative second derivative reverses these signs and gives a local maximum.
“Strict local minimum” means that there is some interval around \(c\) on which \(f(x)>f(c)\) for every \(x\ne c\). A strict local maximum is defined by the reversed inequality. The requirement \(f'(c)=0\) is essential: the theorem concerns stationary points, not every point where a second derivative can be calculated.
Why the Sign of \(f''(c)\) Controls \(f'\)
By definition, \(f''(c)\) is the derivative of \(f'\) at \(c\). Thus, for \(x\ne c\), the difference quotient
is close to \(f''(c)\) when \(x\) is sufficiently close to \(c\). If \(f''(c)>0\), that quotient is positive near \(c\). Because \(f'(c)=0\), the numerator is just \(f'(x)\). The denominator \(x-c\) is negative on the left of \(c\) and positive on the right, so the sign of \(f'(x)\) must differ on the two sides.
This argument uses the derivative of \(f'\) at \(c\), not a claim that \(f''\) has a particular sign throughout an interval. In particular, continuity of \(f''\) is not an additional hypothesis in the theorem above.
Derivation of the Second Derivative Test
We prove the positive case first. Assume \(f'(c)=0\) and \(f''(c)=L>0\). Since the derivative of \(f'\) at \(c\) is \(L\), the definition of derivative guarantees a \(\delta>0\) such that
The inequality implies that the difference quotient is greater than \(L/2\), and therefore is positive. Using \(f'(c)=0\), we have
For \(c-\delta<x<c\), the denominator \(x-c\) is negative, so \(f'(x)<0\). For \(c<x<c+\delta\), the denominator is positive, so \(f'(x)>0\).
It remains to justify that these signs give a strict local minimum. Take any \(x\) with \(c-\delta<x<c\). The Mean Value Theorem applies to \(f\) on \([x,c]\): \(f\) is continuous on that closed interval and differentiable on its interior. There is a point \(\xi\in(x,c)\) such that
Here \(\xi<c\), so \(f'(\xi)<0\), while \(c-x>0\). Hence \(f(c)-f(x)<0\), or \(f(x)>f(c)\). If \(c<x<c+\delta\), the Mean Value Theorem on \([c,x]\) gives a point \(\xi\in(c,x)\) with
because both factors on the right are positive. Thus \(f(x)>f(c)\) on both sides of \(c\), proving that \(c\) is a strict local minimum.
For the negative case, suppose \(f''(c)=L<0\). The derivative definition gives a neighborhood of \(c\) in which the difference quotient is negative. Since \(f'(c)=0\), this means \(f'(x)/(x-c)<0\). Therefore \(f'(x)>0\) to the left of \(c\), and \(f'(x)<0\) to the right. Applying the Mean Value Theorem on either side shows \(f(x)<f(c)\) for nearby \(x\ne c\). So \(c\) is a strict local maximum.
Use \(f'(c)=0\), so the numerator in the difference quotient for \(f'\) becomes \(f'(x)\).
The quotient \(f'(x)/(x-c)\) has the sign of \(f''(c)\) sufficiently near \(c\).
For positive \(f''(c)\), \(f'\) is negative on the left and positive on the right; for negative \(f''(c)\), the signs reverse.
The Mean Value Theorem turns those signs of \(f'\) into strict inequalities comparing nearby values of \(f\) with \(f(c)\).
Worked Examples
Worked Example: A Positive Second Derivative
Classify the stationary point of \(f(x)=x^2-4x+6\).
Differentiate:
Set the first derivative equal to zero to find the stationary point:
The second derivative is \(f''(x)=2\), so \(f''(2)=2>0\). The test gives a strict local minimum at \(x=2\). Its function value is
To check directly, write \(x=2+h\). Then
For every nonzero \(h\), \(h^2>0\), so \(f(2+h)>2=f(2)\). This verifies the strict local minimum.
Worked Example: A Negative Second Derivative
Classify the stationary point of \(g(x)=5-(x+2)^2\).
Differentiate and solve for the stationary point:
The second derivative is \(g''(x)=-2\), so \(g''(-2)=-2<0\). The test gives a strict local maximum at \(x=-2\). The function value is
For a direct check, let \(x=-2+h\). Then
For every nonzero \(h\), \(h^2>0\), so \(5-h^2<5=g(-2)\). Thus nearby values are strictly smaller, as required for a local maximum.
Worked Example: Using the Test with a Quartic
Classify the stationary point at \(x=0\) for \(p(x)=x^4-2x^2\).
Differentiate:
At \(x=0\), \(p'(0)=0\) and \(p''(0)=-4<0\). The second derivative test therefore gives a strict local maximum. The function value is \(p(0)=0\).
We can also check the derivative signs directly near zero. For \(0<|x|<1\), \(x^2-1<0\), and
When \(-1<x<0\), both \(x\) and \(x^2-1\) are negative, so \(p'(x)>0\). When \(0<x<1\), \(x>0\) and \(x^2-1<0\), so \(p'(x)<0\). The function rises toward zero from the left and falls away on the right. As a final check, for \(0<|x|<1\),
because \(x^2>0\) and \(x^2-2<0\). Nearby function values are indeed less than \(p(0)\).
A Common Mistake: Applying the Test Without Stationarity
The sign of \(f''(c)\) alone does not establish an extremum at \(c\). For example, let \(f(x)=x^2\) and examine \(c=1\). Although \(f''(1)=2>0\),
So the hypotheses of the test are not satisfied. In fact, \(f(0.9)=0.81<1=f(1)\), while \(f(1.1)=1.21>1=f(1)\). There is no local minimum or maximum at \(1\). The positive second derivative describes the local behavior of the slope, but the test’s conclusion depends on starting with a stationary point.
A related error is to say only that \(f''(c)>0\), therefore \(f'(x)>0\) near \(c\). The proof shows something different: when \(f'(c)=0\), \(f'(x)\) is negative on the left and positive on the right. The sign change, not positivity on both sides, is what establishes the minimum.
Why the Derivation Matters
The derivation identifies exactly what makes the test work. The second derivative is the rate of change of the first derivative. At a stationary point, the first derivative is zero; a positive rate of change forces it to cross from negative to positive, while a negative rate of change forces it to cross from positive to negative. The Mean Value Theorem then makes the connection between these slope signs and strict comparisons of function values.
It also clarifies the assumptions. The point must be interior, the function must be differentiable around it, and \(f'(c)\) must equal zero. The second derivative at \(c\) must exist and be nonzero for this conclusion. If any required hypothesis is missing, this derivation does not establish a local extremum.
Check Your Understanding
Use the proof and hypotheses above to answer these questions.
- Why does \(f''(c)>0\), together with \(f'(c)=0\), imply that \(f'(x)<0\) just to the left of \(c\)?
- Which theorem turns the signs of \(f'\) on either side of \(c\) into comparisons of function values?
- What local extremum follows when \(f'(c)=0\) and \(f''(c)<0\)?
- Why does \(f''(1)>0\) by itself not establish a local minimum at \(x=1\)?
- In the proof, why is continuity of \(f''\) not required as a separate assumption?
