# Thread: Derivative & Average Rate of Change

1. ## Derivative & Average Rate of Change

Let f(1)= 1/t for t>0. For what value of t is f '(t) equal to the average rate of change on f on [a,b]

A) -sqrt(ab) B) sqrt(ab) C) -1/(sqrt[ab]) D) 1/(sqrt[ab])
E) sqrt [(1/2)(1/b-1/a)]

2. Originally Posted by rawkstar
Let f(1)= 1/t for t>0. For what value of t is f '(t) equal to the average rate of change on f on [a,b]

A) -sqrt(ab) B) sqrt(ab) C) -1/(sqrt[ab]) D) 1/(sqrt[ab])
E) sqrt [(1/2)(1/b-1/a)]

I'm sure you meant $\displaystyle f(t) = \frac{1}{t}$

$\displaystyle f'(t) = -\frac{1}{t^2}$

$\displaystyle f'(t) = \frac{f(b) - f(a)}{b-a}$

$\displaystyle -\frac{1}{t^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a}$

$\displaystyle -\frac{1}{t^2} = \frac{\frac{a-b}{ab}}{b - a}$

finish from here?

3. I don't know how you got from
to

4. Skeeter was merely writing the numerator as a single fraction,
to simplify further in your remaining steps, rawkstar.

(1/b)-(1/a)=(a/a)(1/b)-(b/b)(1/a)=a/(ab)-b/(ab)=(a-b)/(ab)

5. Originally Posted by rawkstar
I don't know how you got from
to
$\displaystyle \frac{1}{b} - \frac{1}{a}$

common denominator is $\displaystyle ab$ ...

$\displaystyle \frac{1}{b} \cdot \frac{a}{a} - \frac{1}{a} \cdot \frac{b}{b}$

$\displaystyle \frac{a}{ab} - \frac{b}{ab}$

put them together ...

$\displaystyle \frac{a-b}{ab}$

6. ok i think I got it, anyways the answer i got was B) sqrt(ab)

7. $\displaystyle As\ t>0,\ 'answer\ A)'\ is\ invalid$