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 (B) (1)/(2) (C) (1)/(10) (D) (1)/(4)

Understand the problem: Understand the problem. We need to find the value of the function f(x) at x=5. However, the function is not defined at x=5 because it would result in a division by zero. Since the function is continuous for all x > 1, we can use the limit process to find f(5).Set up the limit: Set up the limit to find f(5). To find f(5), we need to calculate the limit of f(x) as x approaches 5. lim_(x->5) (sqrt(x-1)-2)/(x-5)Apply L'Hôpital's Rule: Apply L'Hôpital's Rule. Since the limit is of the form 0/0, we can apply L'Hôpital's Rule, which states that if the limit of f(x)/g(x) as x approaches a value c is 0/0 or ±∞/±∞, then the limit is the same as the limit of the derivatives of f(x) and g(x) as x approaches c. lim_(x->5) (sqrt(x-1)-2)/(x-5) = lim_(x->5) (f'(x))/(g'(x))Differentiate the numerator and denominator: Differentiate the numerator and denominator. f'(x) = derivative of sqrt(x-1) - 2 = (1/2)(x-1)^(-1/2) g'(x) = derivative of x - 5 = 1 Now we have: lim_(x->5) (1/2)(x-1)^(-1/2)/(1)Calculate the limit: Calculate the limit. Now we substitute x=5 into the derivative of the numerator: lim_(x->5) (1/2)(x-1)^(-1/2) = (1/2)(5-1)^(-1/2) = (1/2)(4)^(-1/2) = (1/2)(1/2) = 1/4Conclude the solution: Conclude the solution. Since the limit as x approaches 5 of f(x) is 1/4, and the function is continuous for all x > 1, we can conclude that f(5) = 1/4.

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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
(B)
(1)/(2)
(C)
(1)/(10)
(D)
(1)/(4)

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) $1$ $\newline$ B) $\frac{1}{2}$ $\newline$ (C) $\frac{1}{10}$ $\newline$ (D) $\frac{1}{4}$

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Math Problems

Algebra 1

Domain and range of square root functions: equations

Full solution

Q. 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)

1

\newline

\frac{1}{2}

\newline

(C)

\frac{1}{10}

\newline

(D)

\frac{1}{4}

Understand the problem: Understand the problem. $\newline$ We need to find the value of the function $f(x)$ at $x=5$ . However, the function is not defined at $x=5$ because it would result in a division by zero. Since the function is continuous for all x > 1, we can use the limit process to find $f(5)$ .
Set up the limit: Set up the limit to find $f(5)$ . To find $f(5)$ , we need to calculate the limit of $f(x)$ as $x$ approaches $5$ . $\lim_{x\to 5} \frac{\sqrt{x-1}-2}{x-5}$
Apply L'Hôpital's Rule: Apply L'Hôpital's Rule. $\newline$ Since the limit is of the form $0/0$ , we can apply L'Hôpital's Rule, which states that if the limit of $f(x)/g(x)$ as $x$ approaches a value $c$ is $0/0$ or $\pm\infty/\pm\infty$ , then the limit is the same as the limit of the derivatives of $f(x)$ and $g(x)$ as $x$ approaches $c$ . $\newline$ $f(x)/g(x)$ $0$
Differentiate the numerator and denominator: Differentiate the numerator and denominator. $\newline$ $f'(x) =$ derivative of $\sqrt{x-1} - 2 = \frac{1}{2}(x-1)^{-\frac{1}{2}}$ $\newline$ $g'(x) =$ derivative of $x - 5 = 1$ $\newline$ Now we have: $\newline$ $\lim_{x\to5} \frac{\frac{1}{2}(x-1)^{-\frac{1}{2}}}{1}$
Calculate the limit: Calculate the limit. $\newline$ Now we substitute $x=5$ into the derivative of the numerator: $\newline$ $\lim_{x\to 5} \left(\frac{1}{2}\right)(x-1)^{-\frac{1}{2}} = \left(\frac{1}{2}\right)(5-1)^{-\frac{1}{2}} = \left(\frac{1}{2}\right)(4)^{-\frac{1}{2}} = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{4}$
Conclude the solution: Conclude the solution. $\newline$ Since the limit as $x$ approaches $5$ of $f(x)$ is $\frac{1}{4}$ , and the function is continuous for all x > 1, we can conclude that $f(5) = \frac{1}{4}$ .