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Which of the following functions of x is guaranteed by the Extreme Value Theorem to have an absolute maximum on the interval [0, 4]?

a. y = tan x
b. y = tan^-1 x
c. y = x^2 - 16/x^2 + x â 20
d. y = 1/e^x â 1

User Kevmc
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Final answer:

b. y = tan^-1 x is the only function given that is continuous over the entire interval [0, 4], and thus, according to the Extreme Value Theorem, it is guaranteed to have an absolute maximum on this interval.

Step-by-step explanation:

The Extreme Value Theorem guarantees that a continuous function on a closed interval has an absolute maximum and minimum. We will analyze the given functions to determine which one fits this criterion:

  • y = tan x: This function is not continuous on the interval [0, 4] due to vertical asymptotes (tan x is undefined for some values within this interval).
  • y = tan-1 x: This is the inverse tangent function, and it is continuous on all of R, including the interval [0, 4]. Therefore, this function is guaranteed to have an absolute maximum on the interval [0, 4] by the Extreme Value Theorem.
  • y = x2 - 16/x2 + x – 20: This function has a point of discontinuity at x = 0 due to division by zero, so it cannot have an absolute maximum on [0, 4] in accordance with the theorem.
  • y = 1/ex – 1: While ex is continuous, the expression is not properly defined. However, functions composed of exponentials and constants usually are continuous.

Therefore, the correct answer is b. y = tan-1 x, since it is the only function provided that is continuous over the entire interval [0, 4], satisfying the conditions of the Extreme Value Theorem.

User Dan Zuzevich
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