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

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

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

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Video Transcript

Evaluate the instantaneous rate of change of 𝑓 of 𝑥 is equal to two 𝑥 squared plus nine at 𝑥 is equal to negative three.

In this question, we are trying to find the instantaneous rate of change of a function 𝑓 of 𝑥 at 𝑥 is equal to negative three. It is important to note that this is not the same as the average rate of change of the function. We recall 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 divided by 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.

As both ℎ and the interval get smaller and smaller, the slopes of the chords 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. As already mentioned, the given expression 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 interval 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.

We now need to evaluate this limit when 𝑥 is equal to negative three. Let’s begin by finding an expression for 𝑓 of negative three plus ℎ. Using the definition of 𝑓 of 𝑥, we have 𝑓 of negative three plus ℎ is equal to two multiplied by negative three plus ℎ squared plus nine. Squaring negative three plus ℎ gives us nine minus six ℎ plus ℎ squared. And distributing two across the parentheses gives us 18 minus 12ℎ plus two ℎ squared plus nine. Our simplified expression for 𝑓 of negative three plus ℎ is 27 minus 12ℎ plus two ℎ squared.

In a similar way, we can calculate 𝑓 of negative three. This is equal to two multiplied by negative three squared plus nine, which is equal to 27. Substituting these back into our expression, we have the limit as ℎ approaches zero of 27 minus 12ℎ plus two ℎ squared minus 27 all divided by ℎ. The constants cancel as 27 minus 27 is zero. As ℎ approaches zero and is therefore not equal to zero, we can divide the numerator and denominator by ℎ. This gives us the limit as ℎ approaches zero of negative 12 plus two ℎ.

We now have a polynomial in ℎ and can evaluate by direct substitution. When ℎ is equal to zero, we have negative 12 plus two multiplied by zero, which is equal to negative 12. We can therefore conclude that the instantaneous rate of change of the function 𝑓 of 𝑥 is equal to two 𝑥 squared plus nine at 𝑥 equals negative three is negative 12.

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