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Finding maxima and minima using derivatives
In the previous example, we were provided with a graph, and finding the relative extrema was a visual task. However, we will not always be given the graph of a function. What can we do in these cases?
We can use what is known as the first and second derivative tests. These tests are based on Fermat's Theorem about stationary points.
Fermat's Theorem states that, if a function has a relative extremum at \(x=c\) and the function is differentiable at that point, then \(f'(c)=0\).
The points where the derivative of a function is equal to \(0\) are called stationary points. The slope of the function at a stationary point is equal to \(0\).
If we look back at the example of the cubic function, we can observe that the relative maximum and minimum are also points where the slope of the graph is equal to 0. Let's draw tangent lines at the relative extrema!
Fig. 1. Graph of a cubic function with tangent lines at its relative extrema.
There must be a link between derivatives and relative extrema.
Finding maxima and minima using first derivative test
Finding the stationary points is what is known as the First Derivative Test. A stationary point might be a local maximum or local minimum, or it might be neither. To determine this, we use what is known as The Second Derivative Test.
The second derivative test states that if \(f\) is a function with a second derivative, and \(x=c\) is a stationary point, then:
- If \(f''(c)<0\), then \(f(c)\) is a local maximum of \(f\).
- If \(f''(c)>0\), then \(f(c)\) is a local minimum of \(f\).
In words, the Second Derivative Test tells us the following:
If the second derivative at a stationary point is negative, the function has a local maximum at that point.
If the second derivative at a stationary point is positive, the function has a local minimum at that point.
Let's try understanding this process with an example.
Find the local maxima and local minima of the function \(f(x)=2x^3-3x^2-12x+4\), if any.
Solution:
Find the derivative of \(f(x)\) using the Power Rule.
$$f'(x)=6x^2-6x-12$$
Evaluate at a critical point.
$$f'(c)=6c^2-6c-12$$
Apply Fermat's Theorem
$$6c^2-6c-12=0$$
Solve for c by factoring. Start by dividing the equation by \(6\):$$c^2-c-2=0$$
Factor the left-hand side of the equation.$$(c+1)(c-2)=0$$
so
$$c=-1$$
$$c=2$$
Find the second derivative of \(f(x)\):
$$f''(x)=12x-6$$
Evaluate the second derivative at each critical point:
$$f''(-1)=-18$$
$$f''(2)=18$$
Since \(f''(-1)<0\) then, there is a local maximum at \(x=-1\). Its value is \(f(-1)=11\). Since \(f''(2)>0\) then, there is a local minimum at \(x=2\). Its value is \(f(2)=-16\). Let's take a look at the graph of the function to see if this makes any sense.
Fig. 3. Graph of the cubic function signifying its relative extrema.
We have found the precise relative extrema of the function.
It is important to note that if \(f''(c)=0\) the test becomes inconclusive. This might happen because graphs have points with a slope of zero that are not relative extrema. In such cases, it might be worth inspecting the graph of the function.
Find the relative extrema of the function \(f(x)=x^3-2\).
Solution:
Find the derivative of \(f(x)\) using the Power Rule:
$$f'(x)=3x^2$$
Evaluate at a critical point:
$$f'(c)=3c^2$$
Apply Fermat's Theorem:
$$3c^2=0$$
Solve for \(c\):
$$c=0$$
Find the second derivative of \(f\):
$$f''(x)=6x$$
Evaluate the second derivative at the critical point:
$$f''(0)=0$$
Since \(f''(0)=0\) we cannot conclude anything from these tests. Let's now take a look at the graph of the function:
Fig. 4. Graph of a cubic function with no relative extrema.
Note that this function does not have relative extrema, even when we found that its derivative at \(x=0\) is equal to zero. This point is still critical because the slope of the function is equal to \(0\) at that point. Note that the function also does not have a global maximum or a global minimum!
Further information on the function can be obtained by finding more of its derivatives, assuming they exist. This is known as the higher-order derivative test.
Is There a Formula for Finding Maxima and Minima?
Unfortunately, there is no concrete formula for finding the maxima and minima of a function. Locating extrema depends completely on the type of function and the shape of its graph.
Looking at the graph of the function is always a good first step! For example, if the function is a Parabola opening downwards, you can find its global maximum by finding its vertex. If you need to find local maxima and local minima without a graph, you can use the first and second derivative tests that we explored above.
Finding maxima and minima using partial derivatives
Finding maxima and minima using partial derivatives is not covered in the US high school Calculus curriculum. Please see our university section for details on this topic!
Finding Maxima and Minima using Derivatives- Key takeaways
- The absolute maximum or global maximum of a function is the greatest output in its range.
- The absolute minimum or global minimum of a function is the least output in its range.
- A relative maximum or local maximum of a function is an output that is greater than the surrounding outputs.
- A relative minimum or local minimum of a function is an output that is less than the surrounding outputs.
- Minima is the plural of minimum. Maxima is the plural of maximum. Collectively they are known as extrema.
- The first derivative test can be used to find a possible local maximum or local minimum. The second derivative test tells us whether the point is a local maximum or a local minimum.
- The second derivative test is inconclusive if \(f''(c)=0\), in which case taking a look at the graph might be a better idea.
- There is no formula for finding maxima or minima. It is dependent on which function you are studying.
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Frequently Asked Questions about Finding Maxima and Minima Using Derivatives
How do you find maxima and minima by differentiation?
After taking the first derivative, use Fermat’s theorem to find the stationary points. The values for which the derivative is 0 are the extrema.
What does 2nd derivative tell you?
It tells you the nature of the critical points after using the first derivative test.
How to find maxima and minima using derivatives?
By taking the first derivative test and then recognising the nature of the critical points using the second derivative test.
Why is second derivative used in maxima and minima?
To determine the nature of the critical points (or stationary points) obtained from the first derivative test.
What does third derivative tell you?
The third derivative signifies the rate of change of the slope of a curve, i.e. the rate of change of the second derivative.
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