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On the interval [0,b] the number c=0.7868 is the number guaranteed by the Mean value theorem for the function f(x)= cos x. What is the approximate value of b?

(A) 1 (B) 1.25 (C) 1.5 (D) 1.75 (E) 2

1 Answer

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Final answer:

The Mean Value Theorem assures the existence of a point c on the interval [0, b] where the derivative of f(x) is equal to the average rate of change from x=0 to x=b.

For the function f(x)=cos x, at c=0.7868, we find that b is approximately 1.25, as cos(1.25) closely matches the required slope of -0.7071.

Therefore, the correct answer is: option (B) 1.25.

Step-by-step explanation:

The Mean Value Theorem states that for a continuous function on a closed interval, there is at least one point c in the interval where the instantaneous rate of change (the derivative) is the same as the average rate of change over that interval.

For the function f(x) = cos x, the derivative f'(x) = -sin x.

At c = 0.7868, f'(c) should be equal to the average rate of change of the function over the interval [0, b].

To find b, we use the fact that f'(0.7868) = -sin(0.7868) is approximately -0.7071, which is the slope of the secant line over [0, b].

The average rate of change of f(x) from x=0 to x=b is (cos(b) - cos(0)) / (b - 0), which simplifies to:

cos(b) - 1 since cos(0) = 1.

We need to find b such that this expression matches -0.7071.

Trying the given options, we find that for option (B) b = 1.25, cos(1.25) is the closest to matching the value of -0.7071, giving the approximate value of b as 1.25. Thus, the answer is (B) 1.25.

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