Video Transcript
Find the inflection point of the function 𝑓 of 𝑥 is equal to negative five 𝑥 plus 𝑥 minus four raised to the fifth power plus two.
The question gives us a function 𝑓 of 𝑥 which is a polynomial. It wants us to find the inflection point of this function. First, we recall that a point 𝑃 is an inflection point of 𝑓 of 𝑥 if 𝑓 is continuous at the point 𝑃 and the concavity of the curve changes at 𝑃. In other words, we need to find the point 𝑃 where our curve changes from concave upward to concave downward or vice versa. And this actually gives us a method to help us find inflection points. If 𝑥 is equal to 𝑎 at an inflection point of our function 𝑓 of 𝑥, then one of the following must be true. Either the second derivative of our function 𝑓 evaluated at 𝑎 is equal to zero or the second derivative of our function 𝑓 evaluated at 𝑎 does not exist.
And remember, in our case, our function 𝑓 of 𝑥 is a polynomial. This means the second derivative of 𝑓 of 𝑥 with respect to 𝑥 is also a polynomial. And in particular, this means the second derivative of our function 𝑓 of 𝑥 with respect to 𝑥 exists for all real values of 𝑥. So the only possible inflection points of our function 𝑓 of 𝑥 is when the second derivative is equal to zero. It’s worth reiterating at this point that the second derivative of our function being equal to zero does not guarantee that this point is an inflection point. This is just a method to help us find our inflection points.
So to find our inflection points, we’re going to want to find an expression for 𝑓 double prime of 𝑥. There’s a few different ways of differentiating our function for 𝑓 of 𝑥. For example, we could distribute our fifth power over our parentheses by using binomial expansion. And this would work. However, there’s a simpler method. We’ll consider 𝑥 minus four all raised to the fifth power as the composition of two functions. We’ll differentiate this by using the chain rule. We recall the chain rule tells us if 𝑢 is a function of 𝑣 and 𝑣, in turn, is a function of 𝑥, then the derivative of 𝑢 of 𝑣 of 𝑥 is equal to 𝑣 prime of 𝑥 times 𝑢 prime evaluated at 𝑣 of 𝑥.
So to use the chain rule on our function 𝑓 of 𝑥, we’ll start by setting our function 𝑣 of 𝑥 to be the inner function 𝑥 minus four. And this means we’ve rewritten our second term as 𝑣 to the fifth power. So we’ll use 𝑢 of 𝑣 is equal to 𝑣 to the fifth power. Now, to use the chain rule, we need expressions for 𝑢 prime and 𝑣 prime. We can find both of these by using the power rule for differentiation. We get 𝑣 prime of 𝑥 is equal to one and 𝑢 prime of 𝑣 is equal to five 𝑣 to the fourth power. We’re now ready to differentiate our function 𝑓 of 𝑥 term by term. First, the derivative of negative five 𝑥 with respect to 𝑥 is negative five.
Next, we’re differentiating our second term of 𝑥 minus four all raised to the fifth power by using the chain rule. We get 𝑣 prime of 𝑥 times 𝑢 prime evaluated at 𝑣 of 𝑥. Finally, we know the derivative of the constant two is just equal to zero. Substituting in our expressions for 𝑢 prime, 𝑣 prime, and 𝑣 of 𝑥, we get that 𝑓 prime of 𝑥 is equal to negative five plus one multiplied by five times 𝑥 minus four all raised to the fourth power. And of course, we can simplify this to get 𝑓 prime of 𝑥 is equal to negative five plus five times 𝑥 minus four all raised to the fourth power.
And remember, we want to find an expression for our second derivative of 𝑓 of 𝑥. So we need to differentiate this again. And the working out here is very similar. We want to differentiate our second term by using the chain rule, so we’ll set 𝑣 of 𝑥 to be our inner function of 𝑥 minus four. And that means we’ve rewritten this term as five 𝑣 to the fourth power. So we’ll set 𝑢 of 𝑣 to be five 𝑣 to the fourth power. And once again, we can differentiate both of these by using the power rule for differentiation. We get 𝑣 prime of 𝑥 is equal to one and 𝑢 prime of 𝑣 is equal to 20𝑣 cubed.
We’re now ready to find an expression for 𝑓 double prime of 𝑥. We’ll differentiate 𝑓 prime of 𝑥 with respect to 𝑥 term by term. First, the derivative of the constant negative five is equal to zero. Next, we’ll differentiate five times 𝑥 minus four raised to the fourth power by using the chain rule. This gives us 𝑓 double prime of 𝑥 is equal to 𝑣 prime of 𝑥 times 𝑢 prime evaluated at 𝑣 of 𝑥. And if we substitute in our expressions for 𝑢 prime, 𝑣 prime, and 𝑣 of 𝑥, we get that 𝑓 double prime of 𝑥 is equal to 20 times 𝑥 minus four all cubed. Now that we found an expression for 𝑓 double prime of 𝑥, we’re ready to try and find possible inflection points.
Our possible inflection points will be when our second derivative is equal to zero. And in our case, we can see our second derivative will be equal to zero only when 𝑥 minus four is equal to zero. In other words, the only possible inflection point for our function 𝑓 of 𝑥 will be when 𝑥 is equal to four. Remember though, we do need to check that there is in fact an inflection point when 𝑥 is equal to four. We’ll do this by sketching a graph of our second-derivative function. To sketch a graph of our second-derivative function, we notice it’s equal to 20 times 𝑥 minus four all cubed.
To sketch a graph of our second-derivative function, we can notice it’s the same shape as the curve 𝑦 is equal to 𝑥 cubed except we translate it four units right and then stretch it by a factor of 20 vertically. This gives us the following shape where the 𝑥-intercept is equal to four. And we can now see from our sketch, when 𝑥 is greater than four, our curve lies above the 𝑥-axis. So our second derivative is positive. And when 𝑥 is less than four, our curve lies below the 𝑥-axis. So our second derivative is negative. So the concavity of our curve is changing when 𝑥 is equal to four. And therefore, there must be an inflection point in our curve when 𝑥 is equal to four.
The last thing we want to do is find the coordinates of our inflection point. We already know the 𝑥-coordinate of our inflection point will be when 𝑥 is equal to four. To find the 𝑦-coordinate, we’ll substitute 𝑥 is equal to four into our function 𝑓 of 𝑥. This gives us negative five times four plus four minus four raised to the fifth power plus two. And we can then calculate this expression. We see it’s equal to negative 18. And this gives us the coordinates of the only inflection point, four, negative 18.
Therefore, we were able to find the inflection point of the function 𝑓 of 𝑥 is equal to negative five 𝑥 plus 𝑥 minus four raised to the fifth power plus two. The inflection point is four, negative 18.
