how to find average rate of change

Calculus
Tutor: None Selected Time limit: 1 Day

how do you find the average rate of change for 

f(x) = x^2-x-2; [0,3]

Please show work to help me understand

Apr 19th, 2015

The AVERAGE rate of change is  f(3) - f(0) / 3 - 0

f(3) = 3^2 - 3 - 2 = 9 - 3 - 2 = 4

f(0) = 0^2 - 0 - 2 = - 2

Therefore,

4 - (-2) / 3 - 0 = 6 / 3 = 2

Apr 19th, 2015

So formula is f(x) = x^-x-2

The f(x) is the part  I was missing. [0,3] f(x) is 3 or y and 0 =x

It just sseems so easy when shown by you lol:-) tks

Apr 19th, 2015

Oops said that wrong

Apr 19th, 2015

[0,3] is the x interval,  from x = 0 to x = 3

f(x) is the same as y.  You just plug in the x values into the equation to find the y values

then, the average rage is just y2 - y1 / x2 - x1

Apr 19th, 2015

Oh, ok. As long as you know what I mean :)

Apr 19th, 2015

How would it apply to r = -x^2-2;[-1,0]

Apr 19th, 2015

I'm assuming you mean r as the function r(x) = -x^2 -2

Same thing:

You first find r(-1) and r(0), then apply the average rate of change formula

r(0) - r(-1) / 0 -(-1)

r(-1) = -(-1)^2 -2 = -1 - 2 = -3

r(0) = -2

then,  -3 -(-2) / 0 -(-1) = -1/1 = -1

Apr 19th, 2015

I think that's what confusedconfused me the equationequation was setup the way I wrote it. When like this do automatically assume r(x)?

Apr 19th, 2015

r(x) simply means r in terms of x.

All function are in terms of some variable.

f(x) means the function is in terms of the variable x as in f(x) = x^2 - x - 2 or any other equation

r(x) means the same thing

If the function was in terms of time then it would be  r(t) = t^2 + t ...blah blah blah

All functions are in terms of some variable, x or t or p or u or whatever.

Hope that makes sense :)

Apr 19th, 2015

It's marinating now, because for r = problem I got 1 so I'm using formula r(0)-r(-1)/0-(-1)

= -2-(-3)/1

= -2+3/1

= 1/1

= 1

Apr 19th, 2015

Hit enter to quick, where did I go wrong?

Apr 19th, 2015

You're RIGHT, my bad. I reversed the order for r(-1) and r(0)!

I'm glad you're paying attention :)

Apr 19th, 2015

omg tks so much, like I said Im marinating on it, some is sticking lol

Apr 20th, 2015

Yes, it's sticking very well :)

Apr 20th, 2015

do you find the instantaneous rate of change the same way, not working out for me on these 

t = - x^2 = x + 2; -3

g(r) = r^2 – r + 1; 1



Apr 20th, 2015

NO. Instantaneous is VERY different. You have to take the derivative of the function.

t'(x) = -2x + 1

t'(-3) = -2(-3) + 1 = 7

g'(r) = 2r -1

g'(1) = 2(1) = 2

Apr 20th, 2015

Ok I tried it that way first and this is the answer I recieved, but then I am being told it is incorrect and the t'(x) = 5 and g'(r) = 1

I don't understand where I did he formulas wrong

Apr 20th, 2015

The function you wrote was not clear:  t = - x^2 = x + 2 ??

I assumed you mean t = -x^2 + x + 2 because = and + are on the same key on the keyboard

Apr 20th, 2015

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