Video: Finding the Inflection Points of a Function from the Graph of Its Derivative

Using the given graph of the function 𝑓′, at what values of π‘₯ does 𝑓 have an inflection point?

05:19

Video Transcript

Using the given graph of the function 𝑓 prime, at what values of π‘₯ does 𝑓 have an inflection point?

We’re given a graph of the curve of the function 𝑓 prime which is the first derivative of 𝑓 of π‘₯ with respect to π‘₯. We need to use this to determine the values of π‘₯ where our function 𝑓 will have its inflection points. To start, we need to recall what it means for our function 𝑓 to have an inflection point at π‘₯. Our inflection points of the function 𝑓 will be the points where our curve changes concavity. We also require that our function 𝑓 be continuous at this point. We have several different ways of checking this. For example, we know the second derivative of 𝑓 of π‘₯ with respect to π‘₯ will be equal to zero at our points of inflection or they won’t be defined.

However, we need to reiterate just because the second derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to zero or does not exist does not guarantee that these are inflection points. This is just a test for where the possible inflection points are. And in fact, since we’re given a curve 𝑦 is equal to 𝑓 prime of π‘₯, when the second derivative of 𝑓 of π‘₯ is equal to zero, the slope of this curve will be equal to zero. So we can just find the possible inflection points from this graph. For example, we can see the derivative exists for all values of π‘₯.

Next, we just need to find the turning points of our curve. That’s when its slope is equal to zero. We can see this happens at three points β€” when π‘₯ is equal to two, when π‘₯ is equal to four, and when π‘₯ is equal to six. But remember, these are only possible inflection points for our function 𝑓. We still need to check that they are in fact inflection points. And to do this, we’re going to need to go back to our definition of an inflection point. We need to check that our function 𝑓 is continuous at π‘₯ and that our function 𝑓 changes concavity here.

Let’s start with determining whether 𝑓 is continuous at these points. Well, we need to remember a fact about derivatives. We can see that our function 𝑓 is differentiable at each of these three points because they lie on our curve 𝑦 is equal to 𝑓 prime of π‘₯. And we need to remember if a function is differentiable at a point, then it must be continuous at this point. So all three of these points are continuous for our function 𝑓 of π‘₯. So now, all we need to do is check whether the concavity of our curve 𝑦 is equal to 𝑓 of π‘₯ changes at these three points.

To do this, let’s recall what we mean by the concavity of a curve 𝑦 is equal to 𝑓 of π‘₯. Let’s start with what we mean by saying the curve 𝑦 is equal to 𝑓 of π‘₯ is concave upwards. We say that the curve 𝑦 is equal to 𝑓 of π‘₯ is concave upward on an interval if all of its tangent lines lie below the curve on this interval. And we need to remember something interesting. If we check the slope of these tangent lines, we will see that the slope is increasing. So we know that our curve will be concave upward if the slope of our tangent lines is increasing. But remember, 𝑓 prime of π‘₯ actually tells us the slope of our tangent line at π‘₯.

So saying that the slope of our tangent lines is increasing on an interval is the same as saying the curve 𝑦 is equal to 𝑓 prime of π‘₯ is increasing on an interval. So we can find the intervals where 𝑦 is equal to 𝑓 of π‘₯ is concave upward by just looking at the intervals where our curve 𝑦 is equal to 𝑓 prime of π‘₯ is increasing. And in fact, we can do exactly the same thing to find the intervals where 𝑓 of π‘₯ is concave downward.

To start, we need to recall for a curve to be concave downward on an interval is the same as saying its tangent lines lie above the curve. And we also need to recall this is the same as saying the slope of our tangent line is decreasing on this interval. And of course, because 𝑓 prime of π‘₯ measures the slope of our tangent line, saying the slope of our tangent line is decreasing on an interval is the same as saying the curve 𝑦 is equal to 𝑓 prime of π‘₯ is decreasing on that interval. So now we need to ask the question, what does this mean for our three possible inflection points?

Remember, an inflection point will be a point where our curve changes concavity. So we want the points where our curve changes from concave upward to concave downward or from concave downward to concave upward. And as we just explained, one way of doing this is to look for the points where 𝑓 prime changes from an increasing function to a decreasing function or vice versa. So let’s start with the point when π‘₯ is equal to two. We can see when π‘₯ is less than two, 𝑓 prime of π‘₯ is increasing. The curve is going upwards. However, when π‘₯ is greater than two, we can see that our function 𝑓 prime of π‘₯ is decreasing. The curve is going downwards. And this tells us the concavity of the curve 𝑦 is equal to 𝑓 of π‘₯ changes.

Therefore, we’ve shown this is an inflection point. We can then do the same for our other two points. When π‘₯ is less than four, we can see that our curve 𝑓 prime of π‘₯ is decreasing. It’s going downwards. However, when π‘₯ is greater than four, we can see that the curve 𝑓 prime of π‘₯ is increasing. It’s going upwards. Therefore, π‘₯ equals four is also an inflection point for 𝑓 of π‘₯. And we can see exactly the same thing happens at π‘₯ is equal to six. For values less than six, our curve is increasing, and for values greater than six, our curve is decreasing. Therefore, this is also an inflection point for 𝑓 of π‘₯.

And remember, these three points are the only possible inflection points for 𝑓 of π‘₯. And this means we’re done. Therefore, by looking at the intervals where the curve 𝑦 is equal to 𝑓 prime of π‘₯ is increasing and decreasing, we were able to find the inflection points for our function 𝑓 of π‘₯. We were able to show that 𝑓 has inflection points when π‘₯ is equal to two, π‘₯ is equal to four, and when π‘₯ is equal to six.

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