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

Evaluate the rate of change of π(π₯) = (6π₯Β² + 7)/7π₯ at π₯ = 3.

02:52

### Video Transcript

Evaluate the rate of change of π of π₯ equals six π₯ squared plus seven all over seven π₯ at π₯ equals three.

Remember the definition for the instantaneous rate of change of a function at a point π, also known as its derivative at that point, is given by 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 six π₯ squared plus seven all over seven π₯. And weβre looking to find the instantaneous rate of change when π₯ is equal to three. So weβre going to let π be equal to three. We can replace π with three. And we see that the rate of change of the function at π₯ equals three is given by the limit as β approaches zero of π of three plus β minus π of three all over β.

Weβre now going to substitute three plus β and three into our function. π of three plus β is then six lots of three plus β squared plus seven over seven lots of three plus β. And then, π of three is six times three squared plus seven over seven times three. Weβre going to distribute the parentheses in the numerator of the first fraction, and weβre going to evaluate the numerator of the second. When we do, we see that π prime of three, the derivative of our function and therefore the rate of change π₯ equals three, is equal to the limit as β approaches zero of six β squared plus 36β plus 61 over seven times three plus β minus 61 over seven times three all over β.

Now, I specifically didnβt evaluate the denominator of these two fractions. And that was to remind me that each denominator had a factor of seven. So when I subtract by creating a common denominator, I simply need to multiply the numerator and denominator of my first fraction by three and of my second fraction by three plus β. That gives me a common denominator of 21 times three plus β. And then, the numerator is 18β squared plus 108β plus 183 minus 183 minus 61β.

Now, since dividing by β is the same as multiplying by one over β, we can rewrite our denominator as 21β times three plus β. We then spot that 183 minus 183 is zero. And we also see that we can divide both the numerator and the denominator of our fraction by β. That leaves us with the limit as β approaches zero of 18β plus 108 minus 61 over 21 times three plus β. And the numerator here further simplifies to 18β plus 47. At this stage, weβre now ready to use direct substitution to evaluate our limit. Replacing β with zero and we find that π prime of three is 18 times zero plus 47. Thatβs just 47 over 21 times three plus zero. We evaluate the denominator as 63.

And we found the rate of change of our function at π₯ equals three to be 47 over 63.