Let g(x)={[(sqrt(x+20)-4)/(x+4)," for "x = -20","x!=-4],[k," for "x=-4]:} g is continuous for all x -20. What is the value of k ? Choose 1 answer: (A) (1)/(4) (B) -(1)/(8) (C) (1)/(8) () -(1)/(4)

Question

Let  g(x)={[(sqrt(x+20)-4)/(x+4)," for "x >= -20","x!=-4],[k," for "x=-4]:}  g is continuous for all  x > -20. What is the value of  k ? Choose 1 answer: (A)  (1)/(4) (B)  -(1)/(8) (C)  (1)/(8) ()  -(1)/(4)

Bytelearn · Accepted Answer

Step 1: Ensure Continuity: To ensure the function g(x) is continuous at x = -4, the limit of g(x) as x approaches -4 from the left must equal the value of g(x) at x = -4. We need to find the limit of the first piece of the function as x approaches -4.Step 2: Calculate the Limit: We calculate the limit of the first piece of g(x) as x approaches -4: $$ \lim_{{x \to -4}} \frac{\sqrt{x+20}-4}{x+4} $$ To evaluate this limit, we can rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator.Step 3: Rationalize the Numerator: Multiplying by the conjugate, we get: $$ \lim_{{x \to -4}} \frac{\sqrt{x+20}-4}{x+4} \cdot \frac{\sqrt{x+20}+4}{\sqrt{x+20}+4} $$ This simplifies to: $$ \lim_{{x \to -4}} \frac{x+20-16}{(x+4)(\sqrt{x+20}+4)} $$Step 4: Simplify the Expression: Further simplification gives us: $$ \lim_{{x \to -4}} \frac{x+4}{(x+4)(\sqrt{x+20}+4)} $$ Now, we can cancel out the (x+4) term in the numerator and denominator, as long as x is not equal to -4, which is allowed since we are taking a limit as x approaches -4.Step 5: Cancel Out Terms: After canceling out the (x+4) term, we are left with: $$ \lim_{{x \to -4}} \frac{1}{\sqrt{x+20}+4} $$ Now we can directly substitute x = -4 into the expression since it is no longer undefined.Step 6: Substitute x = -4: Substituting x = -4 into the expression, we get: $$ \frac{1}{\sqrt{-4+20}+4} = \frac{1}{\sqrt{16}+4} = \frac{1}{4+4} = \frac{1}{8} $$ So, the limit of g(x) as x approaches -4 from the left is $\frac{1}{8}$.Step 7: Calculate the Limit: Since g(x) must be continuous at x = -4, the value of k must be equal to the limit we just found. Therefore, k = $\frac{1}{8}$.

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