Question

find the value of k that makes the function continuous \( f(x)=\left\{\begin{array}{ll}\sqrt{x-k} & x \geq 7 \\ x-47 & x

Answer 1

The value of
`\( k \)` that makes the function
`\( f(x) = √(x - k) \)` continuous at
`\( x = 7 \) is \( k = 49 \).`

Step-by-step explanation:

A function is continuous at a point if the left-hand limit, right-hand limit, and the function value at that point are all equal. In this case, we have two different expressions for
`\( f(x) \)` based on the intervals
`\( x \geq 7 \)` and
`\( x < 7 \).` For continuity at
`\( x = 7 \),` the expression
`\( √(x - k) \)` must be equal to
`\( x - 47 \)` at
`\( x = 7 \).`

Setting
`\( √(x - k) = x - 47 \)` and solving for
`\( k \) gives \( k = 49 \).` This value makes the function continuous at
`\( x = 7 \)` because both expressions converge to the same value at this point, ensuring the left-hand limit, right-hand limit, and function value are all consistent.

Understanding the conditions for continuity and solving for the parameter
`\( k \)`allows us to identify the specific value that ensures the function is continuous at the specified point. This process is fundamental in calculus and helps ensure that functions behave smoothly without abrupt changes or breaks in their behavior.