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Let 
f(x)={[(x+1)/(sqrt(x+5)-2)," for "x >= -5","x!=-1],[k," for "x=-1]:}

f is continuous for all 
x > -5.
What is the value of 
k ?
Choose 1 answer:
(A) -4
(B) 2
(C) 0
(D) 4

Let f(x)={x+1x+52amp; for x5,x1kamp; for x=1 f(x)=\left\{\begin{array}{ll}\frac{x+1}{\sqrt{x+5}-2} & \text { for } x \geq-5, x \neq-1 \\ k & \text { for } x=-1\end{array}\right. \newlinef f is continuous for all x>-5 .\newlineWhat is the value of k k ?\newlineChoose 11 answer:\newline(A) 4-4\newline(B) 22\newline(C) 00\newline(D) 44

Full solution

Q. Let f(x)={x+1x+52 for x5,x1k for x=1 f(x)=\left\{\begin{array}{ll}\frac{x+1}{\sqrt{x+5}-2} & \text { for } x \geq-5, x \neq-1 \\ k & \text { for } x=-1\end{array}\right. \newlinef f is continuous for all x>5 x>-5 .\newlineWhat is the value of k k ?\newlineChoose 11 answer:\newline(A) 4-4\newline(B) 22\newline(C) 00\newline(D) 44
  1. Define Limit Approach: To find the value of kk that makes f(x)f(x) continuous at x=1x = -1, we need to ensure that the limit of f(x)f(x) as xx approaches 1-1 from the left is equal to the value of f(x)f(x) at x=1x = -1.
  2. Find Limit Expression: First, we will find the limit of the function as xx approaches 1-1 from the left. This means we will use the piece of the function defined for x5x \geq -5 and x1x \neq -1, which is the rational expression x+1x+52\frac{x+1}{\sqrt{x+5}-2}.
  3. Algebraic Manipulation: To find the limit as xx approaches 1-1, we can try direct substitution to see if the expression is defined at x=1x = -1. Substituting x=1x = -1 into the expression gives us (0)/(42)=0/0(0)/(\sqrt{4}-2) = 0/0, which is an indeterminate form. This means we need to use algebraic manipulation to simplify the expression before finding the limit.
  4. Rationalize Denominator: To eliminate the indeterminate form, we can multiply the numerator and denominator by the conjugate of the denominator. The conjugate of x+52\sqrt{x+5}-2 is x+5+2\sqrt{x+5}+2. This will help us rationalize the denominator.
  5. Simplify Expression: Multiplying the numerator and denominator by the conjugate, we get:\newline(x+1)(x+5+2)(x+52)(x+5+2)\frac{(x+1)(\sqrt{x+5}+2)}{(\sqrt{x+5}-2)(\sqrt{x+5}+2)}
  6. Cancel Terms: Simplifying the denominator using the difference of squares, we get:\newline(x+1)(x+5+2)(x+5)4\frac{(x+1)(\sqrt{x+5}+2)}{(x+5) - 4}
  7. Substitute x=1x = -1: Further simplifying the denominator, we get: (x+1)(x+5+2)x+1\frac{(x+1)(\sqrt{x+5}+2)}{x + 1}
  8. Determine Value of kk: Now we can cancel out the (x+1)(x+1) terms in the numerator and denominator, as long as xx is not equal to 1-1 (which is not a problem since we are considering the limit as xx approaches 1-1, not the value at x=1x = -1): x+5+2\sqrt{x+5}+2
  9. Check Answer Choices: Now we can substitute x=1x = -1 into the simplified expression to find the limit: 1+5+2=4+2=2+2=4\sqrt{-1+5}+2 = \sqrt{4}+2 = 2+2 = 4
  10. Check Answer Choices: Now we can substitute x=1x = -1 into the simplified expression to find the limit: 1+5+2=4+2=2+2=4\sqrt{-1+5}+2 = \sqrt{4}+2 = 2+2 = 4Since f(x)f(x) is continuous at x=1x = -1, the limit as xx approaches 1-1 from the left must be equal to the value of f(x)f(x) at x=1x = -1. Therefore, kk must be equal to 44.
  11. Check Answer Choices: Now we can substitute x=1x = -1 into the simplified expression to find the limit: 1+5+2=4+2=2+2=4\sqrt{-1+5}+2 = \sqrt{4}+2 = 2+2 = 4 Since f(x)f(x) is continuous at x=1x = -1, the limit as xx approaches 1-1 from the left must be equal to the value of f(x)f(x) at x=1x = -1. Therefore, kk must be equal to 44. We can now check the answer choices to see which one matches our result. The correct answer is (D) 44.

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