# Question Video: Finding the Rate of Change of a Polynomial Function at a Point

Evaluate the rate of change of 𝑓(𝑥) = 2𝑥² + 9 at 𝑥 = −3.

03:48

### Video Transcript

Evaluate the rate of change of 𝑓 of 𝑥 equals two 𝑥 squared plus nine at 𝑥 equals negative three.

We’re looking for the rate of change of a function and not the average rate of change of the function over an interval, but the instantaneous rate of change of the function at 𝑥 equals negative three.

You might know that the average rate of change of an arbitrary function, like the one graphed over the interval from 𝑥 to 𝑥 plus ℎ, is given by 𝑓 of 𝑥 plus ℎ minus 𝑓 of 𝑥 all over ℎ, that is, the slope of the line segment shown.

Clearly, the slope of the line segment and hence the average rate of change of the function will depend on the size of the interval that we’re averaging over, and hence on the value of ℎ. But as both ℎ and the interval get smaller and smaller, the slopes of the cords get closer and closer to the slope of the tangent line. And it’s the slope of this tangent line that is the instantaneous rate of change of the function at the given value of the input.

The expression 𝑓 of 𝑥 plus ℎ minus 𝑓 of 𝑥 all over ℎ gives the average rate of change of the function on the interval from 𝑥 to 𝑥 plus ℎ. To find the instantaneous rate of change, we take the limit as ℎ approaches zero, making the intervals smaller and smaller and the slopes of those chords closer and closer to the slope of the tangent, that is, the instantaneous rate of change.

Now we just need to evaluate this limit. We’re looking for the instantaneous rate of change at 𝑥 equals negative three. So we substitute negative three for 𝑥. So our numerator becomes 𝑓 of negative three plus ℎ minus 𝑓 of negative three.

We now use the definition of 𝑓 of 𝑥 that we have. We find 𝑓 of negative three by substituting negative three for 𝑥 and the expression we have for 𝑓 of 𝑥. That is, wherever we see an 𝑥, we replace it by negative three.

Now how about the other term, 𝑓 of negative three plus ℎ? How do we find that? Well, in the same way, we just replace 𝑥 in our definition of 𝑓 of 𝑥 by negative three plus ℎ. And so we get two times negative three plus ℎ squared plus nine. And the denominator ℎ completes the expression inside the limit.

Our numerator is quite messy. So let’s see if we can simplify it. We start off with two times negative three plus ℎ squared, which we can expand to get ℎ squared minus six ℎ plus nine. We add nine and subtract 27, which is what 𝑓 of negative three turns out to be, before again dividing by ℎ.

We can simplify further, distributing the two over the terms in the parentheses and then dealing with the other constant terms. We find we have to subtract 18. And so the constant terms cancel. And we’re left with just two ℎ squared minus 12ℎ in the numerator and as ever ℎ in the denominator.

We can see now that the numerator has a factor of ℎ, which we can cancel with the denominator. Canceling the factor, we get the limit of two ℎ minus 12 as ℎ approaches zero, which we can evaluate using direct substitution. It’s just two times zero minus 12, which is negative 12.

The instantaneous rate of change of the function 𝑓 of 𝑥 equals two 𝑥 squared plus nine at 𝑥 equals negative three is, therefore, negative 12. This is the limit of the average rate of change of the function on the interval from 𝑥 to 𝑥 plus ℎ as ℎ approaches zero.