Why Periodicity Matters for Extrema
Sine and cosine repeat their values and derivative patterns. As a result, their local maxima and minima occur in repeating families rather than at just a few isolated inputs. When a problem specifies an interval, the task is to identify which members of those families lie in the interval and classify them using the sign of the derivative.
The sign-change method is useful because it links the graph’s local behavior to increasing and decreasing intervals. For trigonometric functions, periodicity makes the candidate locations predictable, but it does not remove the need to check the specified interval or the derivative signs.
Local Extrema and the Derivative-Sign Test
A function \(f\) has a local maximum at \(x=c\) if there is an open interval around \(c\) on which \(f(c)\geq f(x)\) for every \(x\) in the domain. A local minimum is defined by reversing the inequality. If the domain is a specified closed interval, an endpoint can also be a one-sided local extremum relative to that domain.
For an interior point \(c\), a derivative sign change provides a sufficient test when the function is continuous at \(c\). Sine and cosine, and the transformed trigonometric functions considered here, are continuous everywhere.
A point where \(f'(c)=0\), or where the derivative does not exist, is a critical number when \(c\) is in the domain. A critical number is a candidate for an extremum, not a guarantee of one. In the sine and cosine examples, the derivatives exist everywhere, so we find candidates by solving \(f'(x)=0\).
Periodic Critical Points of Sine and Cosine
The derivative of \(\sin x\) is \(\cos x\). Thus the critical numbers of \(\sin x\) satisfy \(\cos x=0\), giving \(x=\frac{\pi}{2}+k\pi\), where \(k\) is any integer. The signs of the cosine derivative alternate at these points: odd multiples of \(\frac{\pi}{2}\) in the form below give maxima or minima according to their position in the cycle.
The derivative of \(\cos x\) is \(-\sin x\). Its critical numbers satisfy \(\sin x=0\), so \(x=k\pi\). At even multiples of \(\pi\), cosine is \(1\); at odd multiples it is \(-1\). The derivative changes from positive to negative at the former and from negative to positive at the latter.
To use these formulas on a specified interval, solve the appropriate inequalities for the integer \(k\). Checking only one cycle is not enough if the interval covers several cycles.
Why the Sign Change Gives an Extremum
Here is the reasoning behind the derivative-sign test. If \(f'(x)>0\) throughout an interval to the left of \(c\), the Mean Value Theorem implies that \(f\) is increasing there: for \(x_1<x_2<c\), there is a point \(d\) between them such that
Therefore, values to the left of \(c\) increase as the input approaches \(c\). Continuity at \(c\) ensures that these values approach \(f(c)\), so they do not exceed \(f(c)\). If \(f'(x)<0\) to the right, the same argument shows that values decrease as the input moves away from \(c\). Together these facts establish a local maximum. Reversing both signs establishes a local minimum.
Continuity is essential to this conclusion: derivative signs on either side do not, by themselves, determine the value assigned at \(c\). For the trigonometric functions in this tutorial, continuity is assured.
Worked Example: Classifying Sine Extrema on an Interval
Worked Example: Classifying Sine Extrema on an Interval
Find and classify the interior local extrema of \(f(x)=\sin x\) on \([-\pi,2\pi]\). Then check whether either endpoint is a one-sided local extremum relative to this interval.
Differentiate: \(f'(x)=\cos x\). The critical numbers satisfy
Keep only values strictly between \(-\pi\) and \(2\pi\). Substituting integer values of \(k\) gives \(x=-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\). The next values, \(x=-\frac{3\pi}{2}\) and \(x=\frac{5\pi}{2}\), lie outside the interval.
At \(-\frac{\pi}{2}\), the derivative changes from negative to positive, so there is a local minimum. Its value is \(f(-\frac{\pi}{2})=-1\). At \(\frac{\pi}{2}\), the derivative changes from positive to negative, giving a local maximum with value \(1\). At \(\frac{3\pi}{2}\), the derivative changes from negative to positive, giving a local minimum with value \(-1\).
At the left endpoint, \(f(-\pi)=0\), and just to its right the sine values are negative, so \(-\pi\) is a one-sided local maximum relative to the interval. At the right endpoint, \(f(2\pi)=0\), and just to its left the sine values are negative, so \(2\pi\) is also a one-sided local maximum.
Check the interior classifications by substitution: \(\sin(-\frac{\pi}{2})=-1\), \(\sin(\frac{\pi}{2})=1\), and \(\sin(\frac{3\pi}{2})=-1\). These agree with the minimum and maximum values found from the derivative signs.
Worked Example: Classifying Cosine Extrema Across Several Periods
Worked Example: Classifying Cosine Extrema Across Several Periods
Find and classify the interior local extrema of \(g(x)=\cos x\) on \([0,5\pi]\).
The derivative is \(g'(x)=-\sin x\), so critical numbers satisfy \(\sin x=0\), or \(x=k\pi\). The interior of the interval contains \(k=1,2,3,4\), giving \(x=\pi,2\pi,3\pi,4\pi\).
On \((0,\pi)\), \(\sin x>0\), so \(g'(x)<0\); on \((\pi,2\pi)\), \(\sin x<0\), so \(g'(x)>0\). Thus \(x=\pi\) is a local minimum. The signs reverse across \(2\pi\), so it is a local maximum. They alternate at the remaining points: \(3\pi\) is a local minimum and \(4\pi\) is a local maximum.
Substitution checks the values:
These values match the classifications. The endpoints are not included among the interior critical numbers. Relative to the interval, \(0\) is a one-sided local maximum because cosine decreases immediately to its right, and \(5\pi\) is a one-sided local minimum because cosine values immediately to its left are greater than \(-1\).
Worked Example: A Shifted and Stretched Sine Function
Worked Example: A Shifted and Stretched Sine Function
Find and classify the interior local extrema of \(h(x)=2\sin(3x-\frac{\pi}{6})+1\) on \([0,\pi]\).
Differentiate using the chain rule:
Set the derivative equal to zero. The cosine is zero when its angle is \(\frac{\pi}{2}+k\pi\):
For \(0<x<\pi\), the admissible integers are \(k=0,1,2\). The critical numbers are therefore \(\frac{2\pi}{9},\frac{5\pi}{9},\frac{8\pi}{9}\). At the first, the phase is \(\frac{\pi}{2}\); the cosine changes from positive to negative, so \(h'\) changes from positive to negative and the point is a local maximum. At the second, the phase is \(\frac{3\pi}{2}\); the derivative changes from negative to positive, so it is a local minimum. At the third, the phase is \(\frac{5\pi}{2}\); the derivative changes from positive to negative, so it is a local maximum.
Evaluate the function at each point. The sine values at these phases are \(1,-1,1\), respectively, so the corresponding function values are
These values verify the expected high, low, and high values of the stretched sine wave. As an endpoint check, \(h(0)=0\) and \(h(\pi)=2\). The derivative is positive just to the right of \(0\), making \(0\) a one-sided local minimum on this interval; it is negative just to the left of \(\pi\), making \(\pi\) a one-sided local maximum.
A Common Mistake: Forgetting the Repeating Candidates
A frequent error is to find one critical point of sine and stop, even when the interval spans multiple periods. For example, claiming that \(\sin x\) has only one critical number, \(x=\frac{\pi}{2}\), on \([0,4\pi]\) ignores the equation \(\cos x=0\) at every point \(x=\frac{\pi}{2}+k\pi\).
To correct the calculation, impose the interval restriction:
The integer values are \(k=0,1,2,3\), so the critical numbers are \(\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}\). Their derivative signs alternate: maxima occur at \(\frac{\pi}{2}\) and \(\frac{5\pi}{2}\), and minima at \(\frac{3\pi}{2}\) and \(\frac{7\pi}{2}\). The correction is to use the full periodic formula and then restrict its integer index to the interval.
Using the Method Reliably
For a sine, cosine, or shifted and stretched version, a careful solution follows the same sequence:
Include the chain-rule factor for a nontrivial angle such as \(Bx+C\).
Solve \(f'(x)=0\), expressing periodic solutions with an integer index.
Find every indexed solution in the specified interval, not just the first one.
A positive-to-negative change gives a local maximum; a negative-to-positive change gives a local minimum, provided continuity holds at the point.
Substitute the inputs into the original function to find the extremum values, and treat endpoints separately if the domain includes them.
The method matters because periodic graphs can contain many extrema within one interval. Listing all candidates prevents omissions, while the derivative signs justify the classification. A derivative equal to zero alone does not identify a maximum or minimum, and an interior sign-change test should not be used to classify an endpoint without considering the domain’s one-sided behavior.
Check Your Understanding
For each question, use periodic critical-point formulas and derivative signs; distinguish interior points from interval endpoints.
- Find and classify the interior local extrema of \(\sin x\) on \([0,3\pi]\).
- Find the interior local extrema of \(\cos x\) on \([-\pi,3\pi]\), including their function values.
- For \(f(x)=3\sin(2x)\), write the general formulas for its local maxima and local minima.
- Why is a critical number not automatically a local extremum?
- What additional property at \(c\) is needed to use the two-sided derivative-sign test as stated?
