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What is the approximate average rate of change for the function f(x) = 19.5(1.15)^x over the interval x = 5 to x = 10?

User Jakub Vano
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To find the average rate of change of the function f(x) = 19.5(1.15)^x over the interval x = 5 to x = 10, we need to find the slope of the secant line that passes through the points (5, f(5)) and (10, f(10)).

The slope of a secant line passing through two points (x1, y1) and (x2, y2) is given by:

(y2 - y1) / (x2 - x1)

Substituting x = 5 and x = 10 into the function, we get:

f(5) = 19.5(1.15)^5 ≈ 34.3
f(10) = 19.5(1.15)^10 ≈ 87.1

Substituting these values into the slope formula, we get:

(87.1 - 34.3) / (10 - 5) = 52.8 / 5 ≈ 10.56

Therefore, the approximate average rate of change of the function f(x) = 19.5(1.15)^x over the interval x = 5 to x = 10 is approximately 10.56.
User Mamba
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