Let f(t) = 1/t for t>0. For what value of t is f ' (t) equal to the average rate of change of f on [a,b]?
To find the value of t where f'(t) is equal to the average rate of change of f on the interval [a, b], we need to first find the average rate of change of f on [a, b] and then find the value of t that satisfies this condition.
The average rate of change of f on [a, b] is given by the formula:
Average Rate of Change = (f(b) - f(a))/(b - a)
In this case, we are given f(t) = 1/t for t > 0. Therefore, f(b) = 1/b and f(a) = 1/a. Substituting these values into the formula, we get:
Average Rate of Change = (1/b - 1/a)/(b - a) = (a - b)/(ab(b - a))
Now we need to find the value of t for which f'(t) is equal to this average rate of change. To do this, we differentiate f(t) = 1/t with respect to t:
f'(t) = d/dt (1/t) = -1/t^2
Setting f'(t) equal to the average rate of change, we have:
-1/t^2 = (a - b)/(ab(b - a))
To solve this equation for t, we can rearrange it as follows:
t^2 = -ab(b - a)/(a - b)
Since we are looking for a value of t where f'(t) equals the average rate of change, we can ignore the negative sign and solve for t:
t = sqrt(ab(b - a)/(a - b))
Therefore, the value of t at which f'(t) is equal to the average rate of change of f on [a, b] is t = sqrt(ab(b - a)/(a - b)).