Average rate and instantaneous rate?

**nautica17** · August 30th 2009, 11:03 AM

Hi everyone!

I happen to be stumped on this question below.

A function y=f(x) and values of x0 and x1 are given.

y= x^3 , x0= 1 , x1= 2

a. Find the average rate of change of y with respect to x.

b. Find the instantaneous rate of change of y with respect to x at x0.

For a. I got 7 as my answer. I'm curious what everyone else gets. For b. I am stumped and know that you have to find a limit probably. Showing steps would be appreciated as I'd like to see how this is done.

(NOTE: We are not allowed to use derivatives yet, so it all has to be limits and stuff.)

**eXist** · August 30th 2009, 11:49 AM

For instantaneous rate of change:

$\lim_{h\rightarrow0} \frac{f(x + h) - f(x)}{h}$

But now your setting $f(x) = x^3$

So you have something like this:

$\lim_{h\rightarrow0} f(x) = \frac{(x + h)^3 - x^3}{h}$

Hope this helps. By the way, this is the definition of a derivative.

**pedrosorio** · August 30th 2009, 11:55 AM

You probably know that to find the average rate you simple find the change in y and divide it by the change in x:

$\mbox{average rate}=\frac{f(x_1) - f(x_0)}{x_1 - x_0}$

Replacing the values we get:

$\mbox{average rate}=\frac{2^3 - 1^3}{2 - 1}=7$

To get the instantaneous rate you do need to find a limit. It is intuitive that we should take the limit of x1 going to x0, to find the instantaneous rate at x0.

$\mbox{instantaneous rate}=\begin{array}{c}lim\\x_1\rightarrow x_0\end{array} \frac{f(x_1) - f(x_0)}{x_1 - x_0}$

That is:

$\mbox{instantaneous rate}=\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1}$

Can you find this limit?

**nautica17** · August 30th 2009, 12:11 PM

Okay so I got two answers. I don't know which is correct. If I go by what eXist says, then I get 343, but if I go by what pedrosorio says then I get 0. I sort of get the concept but I don't know what to use now. None of the answers make sense to me.

Can anyone nudge me a bit more in the right direction, please?

**pedrosorio** · August 30th 2009, 12:16 PM

You are doing something wrong when you find the limits. Our approaches are equivalent.

The 'h' in eXist's approach is equal to my 'x1 - x0', eXist has made a mistake and put lim x->0, when he meant lim h->0.

**nautica17** · August 30th 2009, 12:31 PM

Okay I got 3 as my answer now. Is this correct or did I mess up again?

**eXist** · August 30th 2009, 12:34 PM

Originally Posted by pedrosorio

You are doing something wrong when you find the limits. Our approaches are equivalent.

The 'h' in eXist's approach is equal to my 'x1 - x0', eXist has made a mistake and put lim x->0, when he meant lim h->0.

Thanks for the correction.

**eXist** · August 30th 2009, 12:36 PM

You should be getting 3 as your answer (Sorry, 0 was a typo, I had 0 up here earlier, so your answer is correct).

If you actually take the derivative (which I know you can't use right now) you get:

$\frac{d}{dx}(x^3) = 3x^2\,dx = 3(1)^2 = 3$

Now, to do it without derivatives, you need to factor the top of this:

$\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1}$

Now to factor two cubes (this is for any case):

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

So what you have is:

$\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1} = \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{(x_1 - 1)(x_1^2 + x_1 + 1)}{x_1 - 1} = \begin{array}{c}lim\\x_1\rightarrow 1\end{array} (x_1^2 + x_1 + 1) = 3 $

Sorry, I had a few typos

. But your answer of 3 is correct. Hopefully all my work is correct now. Sorry again for any confusion.

**nautica17** · August 30th 2009, 01:02 PM

Cool, thank-you guys!

You're the best! Finally, I'm learning something in calculus. Great explanations too.

**Defunkt** · August 30th 2009, 01:18 PM

Originally Posted by eXist

You should be getting 3 as your answer (Sorry, 0 was a typo, I had 0 up here earlier, so your answer is correct).

If you actually take the derivative (which I know you can't use right now) you get:

$\frac{d}{dx}(x^3) = 3x^2\,dx = 3(1)^2 = 3$

Now, to do it without derivatives, you need to factor the top of this:

$\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1}$

Now to factor two cubes (this is for any case):

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

So what you have is:

$\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1} = \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{(x_1 - 1)(x_1^2 + x_1 + 1)}{x_1 - 1} = \begin{array}{c}lim\\x_1\rightarrow 1\end{array} (x_1^2 + x_1 + 1) = 3 $

Sorry, I had a few typos

. But your answer of 3 is correct. Hopefully all my work is correct now. Sorry again for any confusion.

Should use the right notation of $\begin{array}{c}lim\\x\rightarrow 1\end{array} \frac{x^3 - 1^3}{x - 1}$ (for all other limits as well, since $x_1$ is a constant).

:P

**pedrosorio** · August 30th 2009, 03:23 PM

Oops

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