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 weβre looking to find the rate
of change at π₯ equals π₯ one. So weβre going to let π be equal
to π₯ one. Then, weβre interested in finding
the rate of change at π₯ one. So thatβs π prime of π₯ one. Thatβs 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. Itβs simply seven π₯ one squared
plus nine. Letβs substitute these back into
our expression for the rate of change of our function at π₯ one.
Itβs 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. Weβre 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. Itβs 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, letβs 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.