Question

Let f(x) = x^2-x. Find the average rate of change of f over the interval [0,t]. Simplify your answer.

Answer 1

the average rate of change of a function \(f(x)\) over the interval \([a,b]\) is

\[\frac{f(b)-f(a)}{b-a}\]which is just the slope of a line connecting the points \((a,f(a))\) and \((b,f(b))\)

Answer 2

ok this is what I got
average rate of change: f(t)-f(0)/t-0 = t^2-t - 0/t-0 = t^2-t/t

Answer 3

@whpalmer4

Answer 4

that's not simplified...

Answer 5

also, if you are writing equations involving fractions without formatting them, you MUST use parentheses to indicate the proper grouping, such as
(f(t)-f(0))/(t-0) = (t^2-t - 0)/(t-0) = (t^2-t)/t

Answer 6

can you simplify (t^2-t) / 0? I was thinking that is the average rate of change

Answer 7

(t^2-t) / t

Answer 8

hmm, simplified would be just t^2

Answer 9

@whpalmer4

Answer 10

no...

Answer 11

\[\frac{t^2-t}{t} = \frac{t^2}{t} - \frac{t}{t} = \]

Answer 12

t^2/t

Answer 13

@whpalmer4

Answer 14

No, that's not correct. Are you telling me you don't know how to correctly reduce the fractions \[\frac{t^2}{t}\] and \[\frac{t}{t}\]?!?

Answer 15

t^2 / t = t

Answer 16

t^2-1 = t

Answer 17

@whpalmer4

Answer 18

I was not thinking about it when I put my answer... when you divide exponents, you subtract them

Answer 19

I end up with t^2/t - t/t = t

Answer 20

@jtryon still incorrect!

\[\frac{t^2}{t} - \frac{t}{t} =\frac{t^2}{t^1} - \frac{t^1}{t^1} = t^{2-1} - t^{1-1} = t^1-t^0 = t - 1\]!!!

I suggest you do some reviewing of exponents...