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
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
4 - (-2) / 3 - 0 = 6 / 3 = 2
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
Oops said that wrong
[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
Oh, ok. As long as you know what I mean :)
How would it apply to r = -x^2-2;[-1,0]
I'm assuming you mean r as the function r(x) = -x^2 -2
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
I think that's what confusedconfused me the equationequation was setup the way I wrote it. When like this do automatically assume r(x)?
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 :)
It's marinating now, because for r = problem I got 1 so I'm using formula r(0)-r(-1)/0-(-1)
Hit enter to quick, where did I go wrong?
You're RIGHT, my bad. I reversed the order for r(-1) and r(0)!
I'm glad you're paying attention :)
omg tks so much, like I said Im marinating on it, some is sticking lol
Yes, it's sticking very well :)
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
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
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
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
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