Video: Finding the Inflection Points of a Polynomial Function

Find the inflection points of the function 𝑓(π‘₯) = (π‘₯⁴/2) βˆ’ 3π‘₯Β² + 4.

06:57

Video Transcript

Find the inflection points of the function 𝑓 of π‘₯ equals π‘₯ to the fourth power over two minus three π‘₯ squared plus four.

Let’s begin by recalling what we know about the inflection points of a function. They’re the points on the function’s graph where there is a change in the direction of curvature. The function changes from concave up to concave down or vice versa. At the point of inflection itself, the second derivative, which, in this case, will be 𝑓 double prime of π‘₯, is equal to zero. But this can also be true for some local minima and local maxima. So it must also be the case that the second derivative undergoes a change of sign around the π‘₯-value of the inflection point.

This reflects the fact that the second derivative of a function is positive when the function is concave up and is negative when the function is concave down. So if there is a change in the concavity of the function, there will also be a change in the sign of the second derivative. It’s also worth noting that inflection points can occur when the second derivative of the function does not exist. But as our function is a polynomial, all of its derivatives will be polynomials and, therefore, will be defined for all real values of π‘₯. To find the inflection points of this function then, we’re first going to need to find an expression for a second derivative, which we can do by differentiating twice.

We see that our function 𝑓 of π‘₯ is a polynomial. And so in order to find its derivative, we can apply the general power rule of differentiation, which tells us that the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power for real values of π‘Ž and 𝑛 is equal to π‘Žπ‘› multiplied by π‘₯ to the 𝑛 minus oneth power. We multiply it by the original exponent 𝑛, and then we reduce the exponent by one. Before we differentiate, we may find it helpful to recall that the constant four can be written as four multiplied by π‘₯ to the power of zero. Now, we can find the first derivative of our function, applying the power rule of differentiation.

We have a half multiplied by four multiplied by π‘₯ cubed, for the derivative of the first term, minus three multiplied by two π‘₯, for the derivative of the second, plus four multiplied by zero multiplied by π‘₯ to the power of negative one, for the derivative of the third term. But of course, multiplying by zero just gives zero. So we see that the derivative of a constant is just zero. Our first derivative simplifies then to two π‘₯ cubed minus six π‘₯. To find the second derivative, we differentiate again. And you may find it helpful here to think of negative six π‘₯ as negative six π‘₯ to the first power. Applying the power rule of differentiation, we have that 𝑓 double prime of π‘₯ is equal to two multiplied by three multiplied by π‘₯ squared minus six.

Now, that negative six is really negative six multiplied by one multiplied by π‘₯ to the power of zero. But as π‘₯ to the power of zero is just one, this is negative six multiplied by one multiplied by one, which is simply negative six. Our second derivative simplifies then to six π‘₯ squared minus six. Now remember, for the inflection points of a function, we’re looking for where this second derivative is equal to zero first of all. So the next step is to set this expression equal to zero and solve the resulting equation. We have the equation then six π‘₯ squared minus six is equal to zero. We can divide through by six and then add one to each side of the equation, giving π‘₯ squared is equal to one.

The final step is to take the square root of each side of the equation, remembering that we must take plus or minus the square root. So we have π‘₯ equals plus or minus the square root of one, which is simply positive or negative one. So we’ve solved our equation and found the π‘₯-values. But remember, these are the π‘₯-coordinates of possible points of inflection because these points could also be local minima or local maxima. There are two more things we need to do. Firstly, we need to evaluate the function itself for each of these π‘₯-values. Evaluating the function when π‘₯ equals one, first of all, we have one to the fourth power over two minus three multiplied by one squared plus four, which is equal to three over two.

Evaluating the function when π‘₯ equals negative one, we also obtain three over two because our function 𝑓 of π‘₯ is, in fact, an even function. So 𝑓 of π‘₯ is equal to 𝑓 of negative π‘₯. We’ve found the values of the function at each of our possible points of inflection. They’re both equal to three over two. The final step is to evaluate the second derivative a little either side of the π‘₯-values of each possible point of inflection, so that we can determine whether the second derivative undergoes a change of sign. We can choose π‘₯-values of negative two, zero, and positive two, as these are a little either side of each of our possible points of inflection.

Now, we already know that the second derivative is equal to zero when π‘₯ equals negative one and π‘₯ equals one. We now need to evaluate it at each of these other π‘₯-values. When π‘₯ equals negative two, we have 𝑓 double prime of negative two is equal to six multiplied by negative two squared minus six. That’s six multiplied by four, which is 24 minus six, giving 18. And the key point here is that the second derivative is positive when π‘₯ equals negative two. When π‘₯ equals two, the second derivative is again equal to 18. And in fact, our second derivative is also an even function. 𝑓 double prime of π‘₯ is equal to 𝑓 double prime of negative π‘₯.

So we say that when π‘₯ equals two, the second derivative is positive. When π‘₯ equals zero, however, between our possible points of inflection, the second derivative is equal to six multiplied by zero squared minus six which is negative six. And so the second derivative is negative at this π‘₯-value. We, therefore, find that the second derivative does indeed change from positive to zero to negative around the π‘₯-value of negative one. So this does indeed correspond to a point of inflection.

We also find that the second derivative changes from negative to zero to positive around the π‘₯-value of one. So this also corresponds to a point of inflection. So by first finding the points at which the second derivative of our function is equal to zero and then considering whether the second derivative also undergoes a change of sign around these π‘₯-values, we found that our function has two inflection points. The points with coordinates one, three over two and negative one, three over two.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.