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The function f is given by f(x) = x³. The application of the Mean Value Theorem to f on the interval [1,3] guarantees a point in the interval (1,3) at which the slope of the line tangent to the graph of f is equal to which of the following?

(A) 0
(B) ₁₂
(C) ₁₃
(D) ₂₄

User Hckr
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Final answer:

The application of the Mean Value Theorem to f(x) = x³ on the interval [1,3] guarantees a point in the interval (1,3) at which the slope of the tangent line is 13.

Step-by-step explanation:

The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in the open interval (a, b) such that the instantaneous rate of change (the slope of the tangent line) at c is equal to the average rate of change (the slope of the secant line) over the interval [a, b].

In this case, the function f(x)=x³ is continuous and differentiable on the interval [1,3]. Therefore, the Mean Value Theorem guarantees that there is a point c in the interval (1,3) where the slope of the tangent line is equal to the average rate of change between x=1 and x=3.

To find this point, we can first find the average rate of change by evaluating the function at both endpoints:

f(1) = (1)³ = 1

f(3) = (3)³ = 27

The average rate of change is given by (27 - 1) / (3 - 1) = 26 / 2 = 13.

Therefore, the slope of the tangent line at the point c in the interval (1,3) is 13. So, the correct answer is (C) ₁₃.

User Ppi
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