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)]

Please Help

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- Dec 27th 2009, 08:31 AMrawkstarDerivative & 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)]

Please Help - Dec 27th 2009, 08:44 AMskeeter
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? - Dec 27th 2009, 10:48 AMrawkstar
I don't know how you got from

http://www.mathhelpforum.com/math-he...000507f0-1.gif to http://www.mathhelpforum.com/math-he...4660ad30-1.gif - Dec 27th 2009, 11:01 AMArchie Meade
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, 11:04 AMskeeter
- Dec 27th 2009, 12:13 PMrawkstar
ok i think I got it, anyways the answer i got was B) sqrt(ab)

- Dec 29th 2009, 08:26 AMArchie Meade
$\displaystyle As\ t>0,\ 'answer\ A)'\ is\ invalid$