Question

If f(t)=13−5t, what is the average rate of change of f(x) over the interval [1,5]?

a) f′ (x)=−5
b) f′ (x)=13
c) f′ (x)=−5 (The average rate of change is the slope of the secant line between t=1 and t=5.)
d) f′ (x)=0

Answer 1

The average rate of change of the function f(t)=13-5t over the interval [1,5] is −5, which represents the slope of the secant line between t=1 and t=5.

Step-by-step explanation:

If f(t)=13−5t, the average rate of change of f(x) over the interval [1,5] can be calculated using the formula for the slope of the secant line:

Average rate of change = (f(5) - f(1)) / (5 - 1).

Substituting the given values:

= ((13 - 5×5) - (13 - 5×1)) / (5 - 1)

= ((13 - 25) - (13 - 5)) / 4

= ((-12) - 8) / 4

= −20 / 4

= −5

Therefore, the correct answer is: a) f′ (x)=−5.