# Question Video: Finding the Value at Which a Function Has Local Maxima and Minima from the Graph of Its First Derivative Mathematics • Higher Education

The graph of the first derivative πβ² of a continuous function π is shown. At what values of π₯ does π have a local maximum and a local minimum?

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### Video Transcript

The graph of the first derivative π prime of a continuous function π is shown. At what values of π₯ does π have a local maximum and a local minimum?

Now, by looking at the graph, you might think that the local maxima are at π₯ equals two and π₯ equals five and the local minima are at π₯ equals three and π₯ equals seven. But we want to know where the function π has a local minima and maxima. But the graph shown is of π prime, the first derivative of π, and not π itself.

We found that the local maxima of π prime are at π₯ equals two and π₯ equals five and the local minima of π prime are at π₯ equals three and π₯ equals seven. Okay, so itβs not quite as easy as just reading off the local maxima and minima from the graph that weβve been given. But we can use this graph of π prime to find the local maxima and minima of the function π.

First, we use Fermatβs theorem, which states that if a function π has a local maximum or minimum at π and π prime of π exists, then π prime of π is zero. Now, we can see that π prime of π exists for all values of π on the graph. And so Fermatβs theorem it tells us that we should be looking for values of π for which π prime of π is zero. This corresponds to where the graph crosses the π₯-axis. So thatβs at π₯ equals one, π₯ equals six, and π₯ equals eight.

Having found the values of π for which π prime of π is zero, we can read Fermatβs theorem again applying it to our function. If π has a local maximum or minimum at π, then π equals one, π equals six, or π equals eight. Therefore, we can see that there are only three possible values of π, where π could have a local maximum or minimum.

Fermatβs theorem does not tell us, however, for each of these three values of π whether π has a local maximum or a local minimum there or indeed whether it has neither a local maximum nor a local minimum. For this, we need the first derivative test. If π prime changes from a positive to negative at π, then π has a local maximum at π. If π prime changes from negative to positive at π, then π has a local minimum at π. And if π prime doesnβt change sign at π, then π has neither a local maximum nor a local minimum at π.

Letβs apply the first derivative test to our first critical number, π equals one. We know that at π equals one, π prime of π is zero. But what is the sign of π prime on either side of π? Well, looking at the graph, we can see that π prime changes from negative to positive at π equals one. And looking at the first derivative test, this tells us that π has a local minimum at π equals one.

Okay, what about our second critical number, π equals six? Again, we look at the graph. But this time, the sign of π prime is changing from a positive to negative at π equals six. And so there is a local maximum of π at six.

And finally, we consider our last critical number, π equals eight, where we see that π prime changes from negative to positive. And so just like at one, π has a local minimum at eight. So hereβs our answer then: π has a local maximum point at π₯ equals six and local minimum points at π₯ equals one and π₯ equals eight.

We used Fermatβs theorem to show that we could only have a local maximum or minimum of π at the critical numbers one, six, and eight, where the derivative of π was zero. And then, we applied the first derivative test at each of those critical numbers to decide whether there was a local maximum, a local minimum, or neither there.

Itβs important that you can not only apply Fermatβs theorem and the first derivative test, but that you can also remember them and understand intuitively why they are true. Near a local maximum of a function, the graph of the function looks like this, with the maximum itself appearing at the very top of the hill. We can see that the tangent to the graph at this local maximum is horizontal and so has a slope of zero. And so the derivative of the function at that point must be zero.

So if the local maximum on the graph is at the point π π of π, then π prime of π is zero. The same is true for a local minimum. The tangent to the graph at that point is horizontal and so has a slope of zero. And so the derivative of the function at that point is zero.

Looking at the sketches, it looks reasonable then that Fermatβs theorem is true. If π has a local maximum or minimum at π, then π prime of π, the derivative of π at π, is zero.

Well, how about our first derivative test? Letβs look at the sketch of our local maximum. The tangent to the graph for a value of π₯ less than π has a positive slope, whereas for a value of π₯ just above π, it has negative slope. So π prime changes from positive to negative at π, whereas at a local minimum, the sign of π prime changes from negative to positive.

So if you ever forget which way around the sign of π prime has to change for that to be a local maximum or a local minimum using the first derivative test, then donβt panic, just draw a sketch as we have done now. The key thing to remember is that the derivative of a function π at a point π₯ is the slope of the tangent to the graph of the function π at that point.