- #1

championnn

- 1

- 0

this is what I did

2-2(1)^2/2-1 = 0/1= 0

is that right?

thanks

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- Thread starter championnn
- Start date

- #1

championnn

- 1

- 0

this is what I did

2-2(1)^2/2-1 = 0/1= 0

is that right?

thanks

- #2

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,253

- 63

Average rate of change would be something like:

[tex] \frac {F(x_1) - F(x_2)} { x_1 -x_2} [/tex]

Your F(x) :

[tex] F(x) = x - 2x^2 [/tex]

Keep in mind, that x is common in that expression, the first x must equal the 2nd x.

- #3

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

this is what I did

2-2(1)^2/2-1 = 0/1= 0

is that right?

thanks

What you wrote, even if the calculations are correct, is meaningless.

First, I suspect you MEAN (2- 2(1)^2)/2- 1, not what you wrote.

In any case, you need to write "the average rate of change is ... " so someone reading this will know what you are doing.

What is the value of the function when x= 1?

What is the value of the function when x= 2?

What is the change in the value?

What is the average rate of change?

It strongly urge you to get into the habit of writing your work one step at a time like that. It will make it clearer to others what you are doing and, hopefully, make more sense to

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