Question 9 (1 point)
[tex]L(x)=\left\{\begin{array}{l}x+5, \text { if } x \neq-1 \\ 2 k x+2, \text { if } x=-1\end{array}\right.[/tex]. Determine [tex]k[/tex] so that [tex]L(x)[/tex] is continuous.
a) [tex]k=-3[/tex]
b) [tex]k=-5[/tex]
c) [tex]k=-1[/tex]
d) [tex]k=6[/tex]
Answer (2)
Find the limit of L ( x ) as x approaches − 1 : lim x → − 1 ( x + 5 ) = 4 .
Evaluate L ( − 1 ) = 2 k ( − 1 ) + 2 = − 2 k + 2 .
Set the limit equal to the function value at x = − 1 : 4 = − 2 k + 2 .
Solve for k : k = − 1 .
Explanation
Problem Analysis We are given a piecewise function L ( x ) and we need to find the value of k that makes the function continuous. For L ( x ) to be continuous at x = − 1 , the limit of L ( x ) as x approaches − 1 must equal the value of L ( − 1 ) .
Finding the Limit First, let's find the limit of L ( x ) as x approaches − 1 . Since we are looking at the limit as x approaches − 1 , we use the definition of L ( x ) for x = − 1 , which is L ( x ) = x + 5 . So, we have x → − 1 lim L ( x ) = x → − 1 lim ( x + 5 ) Substituting x = − 1 into the expression x + 5 , we get x → − 1 lim ( x + 5 ) = − 1 + 5 = 4 So, the limit of L ( x ) as x approaches − 1 is 4 .
Finding L(-1) Next, we need to find the value of L ( − 1 ) . According to the definition of the function, when x = − 1 , L ( x ) = 2 k x + 2 . Therefore, L ( − 1 ) = 2 k ( − 1 ) + 2 = − 2 k + 2 So, L ( − 1 ) = − 2 k + 2 .
Solving for k For L ( x ) to be continuous at x = − 1 , we must have x → − 1 lim L ( x ) = L ( − 1 ) Substituting the values we found, we get 4 = − 2 k + 2 Now, we solve for k :
4 − 2 = − 2 k 2 = − 2 k k = − 2 2 = − 1 Therefore, k = − 1 .
Final Answer Thus, the value of k that makes L ( x ) continuous is k = − 1 .
Examples
In electrical engineering, understanding continuity is crucial when designing circuits. For example, if L ( x ) represents the voltage at a certain point in a circuit depending on a parameter x , ensuring L ( x ) is continuous means that small changes in x won't cause sudden, drastic changes in voltage, which could damage the circuit. Determining the correct value of k ensures the voltage function remains stable and predictable, preventing potential failures.
To make the function L ( x ) continuous at x = − 1 , we find that k must be − 1 . This is derived by setting the limit of L ( x ) as x approaches -1 equal to the value of the function at x = − 1 . Therefore, the answer is k = − 1 .
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