Let g(x)=(x-5)/(sqrt(x-4)-1) when x!=5. g is continuous for all x 4. Find g(5). Choose 1 answer: (A) 5 (B) 10 (C) 2 (D) 8

Question

Let  g(x)=(x-5)/(sqrt(x-4)-1) when  x!=5.  g is continuous for all  x > 4. Find  g(5). Choose 1 answer: (A) 5 (B) 10 (C) 2 (D) 8

Bytelearn · Accepted Answer

Given function: We are given the function g(x) = (x - 5) / (sqrt(x - 4) - 1) and we need to find the value of g(5). Direct substitution of x = 5 into the function would result in a 0/0 indeterminate form, so we need to manipulate the function to remove the discontinuity at x = 5.Rationalizing the denominator: To remove the discontinuity, we can rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate of sqrt(x - 4) - 1 is sqrt(x - 4) + 1.Simplifying the expression: Multiply the numerator and denominator by the conjugate to simplify the expression: g(x) = [(x - 5) * (sqrt(x - 4) + 1)] / [(sqrt(x - 4) - 1) * (sqrt(x - 4) + 1)].Further simplifying the denominator: Simplify the denominator using the difference of squares formula (a - b)(a + b) = a^2 - b^2: g(x) = [(x - 5) * (sqrt(x - 4) + 1)] / [(x - 4) - 1].Cancellation of terms: Further simplify the denominator: g(x) = [(x - 5) * (sqrt(x - 4) + 1)] / (x - 5).Substituting x = 5: Now, we can cancel out the (x - 5) term in the numerator and the denominator, as long as x is not equal to 5. However, since we are interested in the limit as x approaches 5, this cancellation is valid for the purpose of finding g(5): g(x) = sqrt(x - 4) + 1.Calculating the result: Substitute x = 5 into the simplified expression to find g(5): g(5) = sqrt(5 - 4) + 1.Calculate the square root and the sum: g(5) = sqrt(1) + 1 = 1 + 1 = 2.

Let $g(x)=\frac{x-5}{\sqrt{x-4}-1}$ when $x \neq 5$ . $\newline$ $g$ is continuous for all x>4 . $\newline$ Find $g(5)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $5$ $\newline$ (B) $10$ $\newline$ (C) $2$ $\newline$ (D) $8$

Full solution

More problems from Conjugate root theorems

Related topics

Let g(x)=x−5x−4−1 g(x)=\frac{x-5}{\sqrt{x-4}-1} g(x)=x−4​−1x−5​ when x≠5 x \neq 5 x=5.\newlineg g g is continuous for all x>4 .\newlineFind g(5) g(5) g(5).\newlineChoose 111 answer:\newline(A) 555\newline(B) 101010\newline(C) 222\newline(D) 888

Let $g(x)=\frac{x-5}{\sqrt{x-4}-1}$ when $x \neq 5$ . $\newline$ $g$ is continuous for all x>4 . $\newline$ Find $g(5)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $5$ $\newline$ (B) $10$ $\newline$ (C) $2$ $\newline$ (D) $8$