Let f(x)=(sqrt(x-1)-2)/(x-5) when x!=5. f is continuous for all x 1. Find f(5). Choose 1 answer: (A) (1)/(2) (B) (1)/(4) (C) 1 (D) (1)/(10)

Question

Let  f(x)=(sqrt(x-1)-2)/(x-5) when  x!=5.  f is continuous for all  x > 1. Find  f(5). Choose 1 answer: (A)  (1)/(2) (B)  (1)/(4) (C) 1 (D)  (1)/(10)

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Recognize Function Undefined: First, we need to recognize that the function f(x) is not defined at x=5 because it would result in a division by zero. However, we are asked to find f(5), which suggests we need to find the limit of f(x) as x approaches 5. This is because the function is continuous for all x > 1 except at x=5.Find Limit as x Approaches 5: To find the limit of f(x) as x approaches 5, we can use the fact that if the limit exists, the function can be made continuous by defining f(5) to be that limit. We will apply L'Hôpital's Rule because the current form of the function results in an indeterminate form of 0/0 when x=5.Apply L'Hôpital's Rule: L'Hôpital's Rule states that if the limit as x approaches c of f(x)/g(x) is an indeterminate form 0/0 or ∞/∞, then the limit is the same as the limit of the derivatives of the numerator and the denominator, provided that the limit of the derivatives exists. So we will find the derivatives of the numerator and the denominator of f(x) with respect to x.Derivative of Numerator: The numerator of f(x) is sqrt(x-1) - 2. The derivative of sqrt(x-1) with respect to x is (1/2)(x-1)^(-1/2), and the derivative of -2 is 0. So the derivative of the numerator is (1/2)(x-1)^(-1/2).Derivative of Denominator: The denominator of f(x) is x - 5. The derivative of x - 5 with respect to x is 1.Apply L'Hôpital's Rule Again: Now we apply L'Hôpital's Rule by taking the limit of the derivatives as x approaches 5. The limit of the numerator's derivative is (1/2)(5-1)^(-1/2) = (1/2)(4)^(-1/2) = (1/2)(1/2) = 1/4.Calculate Final Limit: The limit of the denominator's derivative as x approaches 5 is simply 1.Determine f(5): The limit of f(x) as x approaches 5 is therefore the limit of the numerator's derivative divided by the limit of the denominator's derivative, which is (1/4) / 1 = 1/4.Since the limit of f(x) as x approaches 5 is 1/4, we can say that f(5) is 1/4 for the purposes of making the function continuous at x=5. This corresponds to answer choice (B).

Let $f(x)=\frac{\sqrt{x-1}-2}{x-5}$ when $x \neq 5$ . $\newline$ $f$ is continuous for all x>1 . $\newline$ Find $f(5)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2}$ $\newline$ (B) $\frac{1}{4}$ $\newline$ (C) $1$ $\newline$ (D) $\frac{1}{10}$

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