Video: Evaluating the Rate of Change of a Polynomial Function at a Point

Evaluate the rate of change of 𝑓(π‘₯) = 7π‘₯Β² + 9 at π‘₯ = π‘₯₁.

02:33

Video Transcript

Evaluate the rate of change of 𝑓 of π‘₯ equals seven π‘₯ squared plus nine at π‘₯ equals π‘₯ one.

Remember, the definition for the rate of change of a function or its derivative at some point π‘₯ equals π‘Ž is the limit as β„Ž approaches zero of 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž all over β„Ž. In this question, 𝑓 of π‘₯ is equal to seven π‘₯ squared plus nine. And we’re looking to find the rate of change at π‘₯ equals π‘₯ one. So we’re going to let π‘Ž be equal to π‘₯ one. Then, we’re interested in finding the rate of change at π‘₯ one. So that’s 𝑓 prime of π‘₯ one. That’s therefore equal to the limit as β„Ž approaches zero of 𝑓 of π‘₯ one plus β„Ž minus 𝑓 of π‘₯ one all over β„Ž.

Our next job is to substitute π‘₯ one plus β„Ž and π‘₯ one into our original function. So 𝑓 of π‘₯ one plus β„Ž is seven times π‘₯ one plus β„Ž squared plus nine. We distribute π‘₯ one plus β„Ž all squared. And we get π‘₯ one squared plus two π‘₯ one β„Ž plus β„Ž squared. And then, we distribute again and we get seven π‘₯ one squared plus 14π‘₯ one β„Ž plus seven β„Ž squared plus nine. 𝑓 of π‘₯ one is a little simpler. It’s simply seven π‘₯ one squared plus nine. Let’s substitute these back into our expression for the rate of change of our function at π‘₯ one.

It’s the limit shown. And, of course, we can distribute the parentheses. And our final two terms become negative seven π‘₯ one squared minus nine. And this is great because seven π‘₯ one squared minus seven π‘₯ one squared is zero and nine minus nine is also zero. And 𝑓 prime of π‘₯ one is, therefore, the limit as β„Ž approaches zero of 14π‘₯ one β„Ž plus seven β„Ž squared over β„Ž. We’re still not quite ready to perform direct substitution. But we can divide both of our terms on the numerator by β„Ž. And now, the rate of change is the limit as β„Ž approaches zero of 14π‘₯ one plus seven β„Ž. And now, we can perform direct substitution. It’s 14π‘₯ one plus seven times zero, which is, of course, just 14π‘₯ one. The rate of change of 𝑓 of π‘₯ equals seven π‘₯ squared plus nine at π‘₯ equals π‘₯ one is 14π‘₯ one.

In this example, we ended up finding a general equation for the rate of change of the function. We could use this to find the particular rate of change at any point, given a value for π‘₯ one. For instance, let’s say we wanted to find the rate of change of the function at the point where π‘₯ equals two. We let π‘₯ one be equal to two. And the rate of change becomes 14 times two, which is 28.

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