Video: Finding All Inflection Points of the Curve of a Polynomial Function

Find all the inflection points of 𝑓(π‘₯) = π‘₯⁴ βˆ’ 2π‘₯Β² + 5.

04:46

Video Transcript

Find all the inflection points of the function 𝑓 of π‘₯ is equal to π‘₯ to the fourth power minus two π‘₯ squared plus five.

The question gives us a function 𝑓 of π‘₯ which is a polynomial, and it wants us to find all the inflection points of this function. And we recall that we call 𝑝 an inflection point of the function 𝑓 of π‘₯ if 𝑓 of π‘₯ is continuous at 𝑝 and the concavity of our function 𝑓 of π‘₯ changes at 𝑝. And we can tell the concavity of a function by checking the sign of the second derivative of that function. So because our function 𝑓 of π‘₯ is a polynomial, we can find our inflection points by finding where our second derivative changes sign. So we need to find an expression for our second derivative of 𝑓 of π‘₯.

To do this, we need to differentiate 𝑓 of π‘₯ twice. Since 𝑓 of π‘₯ is a polynomial, we can differentiate this term by term by using the power rule for differentiation. We need to multiply by the exponent of π‘₯ and then reduce this exponent by one. This gives us 𝑓 prime of π‘₯ is equal to four π‘₯ cubed minus four π‘₯. Now, to find an expression for 𝑓 double prime of π‘₯, we need to differentiate 𝑓 prime of π‘₯ with respect to π‘₯. Again, 𝑓 prime of π‘₯ is a polynomial, so we can do this term by term by using the power rule for differentiation. This time we get 𝑓 double prime of π‘₯ is equal to 12π‘₯ squared minus four. So we want to use all of this information to find all of the inflection points of our function 𝑓 of π‘₯.

First, remember, our function 𝑓 of π‘₯ must be continuous at any inflection point 𝑝. And our function 𝑓 of π‘₯ is a polynomial, so it’s continuous for all real values of π‘₯. So we’ll find our inflection points by just checking where the sign of 𝑓 double prime of π‘₯ changes. There’s a few different ways of doing this. We’re going to do this by sketching a graph of 𝑓 double prime of π‘₯. First, let’s factor our quadratic. We’ll start by taking out a factor of four. This gives us four times three π‘₯ squared minus one. And now we can see that we’re left with four times a difference between squares.

And remember, we can factor the difference between two squares, π‘Ž squared minus 𝑏 squared, as π‘Ž minus 𝑏 times π‘Ž plus 𝑏. So by factoring this difference between squares where π‘Ž is root three π‘₯ and 𝑏 is one, we get that 𝑓 double prime of π‘₯ is equal to four times root three π‘₯ minus one times root three π‘₯ plus one. We can now find the roots of our quadratic. Solving each factor is equal to zero, we get π‘₯ is equal to one over root three or π‘₯ is equal to negative one over root three. And we can rationalize this denominator by multiplying the numerator and the denominator by root three. We get that one over root three is equal to root three over three.

We now have all the information we need to sketch a graph of 𝑓 double prime of π‘₯. First, it’s a quadratic with two unique roots, one at π‘₯ is equal to root three over three and the other at π‘₯ is equal to negative root three over three. Remember, 𝑓 double prime of π‘₯ is equal to 12π‘₯ squared minus four. So the leading term has a positive coefficient, so it will have a similar shape to 𝑦 is equal to π‘₯ squared. So our curve 𝑦 is equal to 𝑓 double prime of π‘₯ will be a parabola with a positive leading coefficient.

We now want to see where 𝑓 double prime of π‘₯ changes sign. We can see when π‘₯ is less than negative root three over three, 𝑓 double prime of π‘₯ is positive. And when π‘₯ is just a little bit bigger than negative root three over three, 𝑓 double prime of π‘₯ is negative. So 𝑓 double prime of π‘₯ changes sign at π‘₯ is equal to negative root three over three. Therefore, this is an inflection point. We can say something similar at π‘₯ is equal to root three over three. When π‘₯ is greater than root three over three, our function 𝑓 double prime of π‘₯ is positive. However, when π‘₯ is a little bit smaller than root three over three, we can see that 𝑓 double prime of π‘₯ is negative. So we’ve shown when π‘₯ is equal to positive or negative root three over three, we have inflection points.

The last thing we need to do is find coordinates for our inflection points. We’ll start by substituting π‘₯ is equal to root three over three into our expression for 𝑓 of π‘₯. Doing this, we get root three over three to the fourth power minus two times root three over three squared plus five. Simplifying this expression, we get one-ninth minus two-thirds plus five, which we can calculate to give us 40 divided by nine. So we’ve shown the 𝑦-coordinate of our inflection point at π‘₯ is equal to root three over three is 40 divided by nine.

Let’s now find the 𝑦-coordinate of our other inflection point. We substitute π‘₯ is equal to negative root three over three into our function 𝑓 of π‘₯. We get negative root three over three to the fourth power minus two times negative root three over three squared plus five. And if we calculate this, we can see we also get 40 divided by nine. Therefore, we’ve shown the function 𝑓 of π‘₯ is equal to π‘₯ to the fourth power minus two π‘₯ squared plus five has two inflection points, one of the coordinates root three over three, 40 over nine, and the other of coordinates negative root three over three, 40 over nine.

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