# Derivative & Average Rate of Change

• Dec 27th 2009, 09:31 AM
rawkstar
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)]

• Dec 27th 2009, 09:44 AM
skeeter
Quote:

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 $f(t) = \frac{1}{t}$

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

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

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

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

finish from here?
• Dec 27th 2009, 11:48 AM
rawkstar
• Dec 27th 2009, 12:01 PM
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)
• Dec 27th 2009, 12:04 PM
skeeter
Quote:
$\frac{1}{b} - \frac{1}{a}$

common denominator is $ab$ ...

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

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

put them together ...

$\frac{a-b}{ab}$
• Dec 27th 2009, 01:13 PM
rawkstar
ok i think I got it, anyways the answer i got was B) sqrt(ab)
• Dec 29th 2009, 09:26 AM
$As\ t>0,\ 'answer\ A)'\ is\ invalid$