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For $y=x^2+1$
Find the average rate of change of $y$ with respect to $x$ over the interval [3,5].

I did it $5=3^2+1$ =5
but the answer is 8
I think there must be formula for average rate of change but doesn't give the formula in the book.

Find the instantaneous rate of change of y with respect to $x$ at the point $x=-4$

I have no idea to start this question, the answer is =-8

Thanks in advance.

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    Just a general piece of advice, Sb Sangpi --- unless you have decided you do not want to learn this material (which means that you have decided that you are not going to pursue math or physics beyond this class), you should really, really not just look up a formula in the book when you run across an exercise. Instead, you should do your best to try to understand *why that formula is relevant to this exercise.* Where does the formula come from? What is it communicating? Just looking for a formula will dull your intellect as well as make the work boring instead of interesting.2012-04-22

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You do not need to differentiate to find the average rate of change. In fact, that would rather over-complicate things.

More simply: The average rate of change = total change / length on x-axis (this comes from the definition of mean average)

When $x = 3, y = (3)^2 + 1 = 10$ When $x = 5, y = (5)^2 + 1 = 26$

Thus, the total change is $26-10 = 16$

The length on the x-axis is 2. And the average rate of change is $16/2 = 8$

To find the rate of change at a single point (instantaneous rate of change), you do need to differentiate first, then substitute $x=-4$ to find the gradient at that particular point.

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    yes :) but by the Fundamental Theorem of Calculus, that simplifies to my method... ;)2012-04-22