MHB Average rate of change in the function from x1 to x2

AI Thread Summary
To find the average rate of change of the function f(x) = x^2 + 12x - 4 from x1 to x2, the formula used is (f(x2) - f(x1)) / (x2 - x1). The discussion clarifies that specific values for x1 and x2 were provided, which the original poster initially overlooked. The importance of using general variables in the formula was emphasized, allowing for flexibility in calculations. The poster expressed relief upon realizing their mistake and appreciated the guidance received. Understanding the average rate of change is essential for precalculus concepts.
Taryn1
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So this is my first question here, and I hope I'm doing it right!

My question is basically this:

Find the average rate of change of the function from x1 to x2.

f(x) = x^2 + 12x -4

I'm also new to precalc, so please don't blame me if this is a really easy question! It doesn't seem to make sense to me, I think I overthink stuff sometimes. Do I just pick two x-values and then find the difference between the results?

(x^2 means x to the power of 2 or x squared)
 
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Oh my gosh...title typo. Meant average, obviously.
 
Taryn said:
So this is my first question here, and I hope I'm doing it right!

My question is basically this:

Find the average rate of change of the function from x1 to x2.

f(x) = x^2 + 12x -4

I'm also new to precalc, so please don't blame me if this is a really easy question! It doesn't seem to make sense to me, I think I overthink stuff sometimes. Do I just pick two x-values and then find the difference between the results?

(x^2 means x to the power of 2 or x squared)

Hi Taryn, (Wave)

Welcome to MHB! I fixed the title for you. No worries - typos happen!

Ok so in general to find the average rate of change between two points, $a$ and $b$, we use:
$$\text{Average change = } \frac{f(b)-f(a)}{b-a}$$

So you were only asked about the general variables, $x_1$ and $x_2$? Were these given any particular values?
 
Yeah, it was only the general variables x1 and x2, no particular values. So I just make up my own?

And thanks for the help!
 
Taryn said:
Yeah, it was only the general variables x1 and x2, no particular values. So I just make up my own?

And thanks for the help!

I would leave it in general terms of $x_1$ and $x_2$ then. What do you get when you plug in the following?
$$\frac{f(x_2)-f(x_1)}{x_2-x_1}$$
 
Wow, I just realized I had made a super dumb mistake! Thanks for your help, Jameson. I've got it now - they gave me values for {x}_{1} and {x}_{2}. lol!
 
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