# Question Video: Evaluating the Rate of Change of a Polynomial Function at a Point Mathematics • Higher Education

Evaluate the rate of change of π(π₯) = 7π₯Β² + 9 at π₯ = π₯β.

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### Video Transcript

Evaluate the rate of change of π 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 is the limit as β approaches zero of π of π plus β minus π of π all over β. In this question, π of π₯ is equal to seven π₯ squared plus nine. And weβre looking to find the rate of change at π₯ equals π₯ one. So weβre going to let π be equal to π₯ one. Then, weβre interested in finding the rate of change at π₯ one. So thatβs π prime of π₯ one. Thatβs therefore equal to the limit as β approaches zero of π of π₯ one plus β minus π of π₯ one all over β.

Our next job is to substitute π₯ one plus β and π₯ one into our original function. So π of π₯ one plus β is seven times π₯ one plus β squared plus nine. We distribute π₯ one plus β all squared. And we get π₯ one squared plus two π₯ one β plus β squared. And then, we distribute again and we get seven π₯ one squared plus 14π₯ one β plus seven β squared plus nine. π of π₯ one is a little simpler. Itβs simply seven π₯ one squared plus nine. Letβs substitute these back into our expression for the rate of change of our function at π₯ one.

Itβ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 π prime of π₯ one is, therefore, the limit as β approaches zero of 14π₯ one β plus seven β squared over β. Weβre still not quite ready to perform direct substitution. But we can divide both of our terms on the numerator by β. And now, the rate of change is the limit as β approaches zero of 14π₯ one plus seven β. And now, we can perform direct substitution. Itβs 14π₯ one plus seven times zero, which is, of course, just 14π₯ one. The rate of change of π 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, letβ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.