Question

Determine the value for k that will make the function continuous at$c=4$ $f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-9}{x+3}ifx\le 4\\ kx+5ifx>4\end{array}\right.$

Short answer

The value of$k=$$-1$

Step-by-step solution

Step 1. Value of the function at c=4

The function where$c=4$is

$f\left(x\right)=\frac{x^2-9}{x+3}$

Substituting the values,

$f\left(4\right)=\frac{4^2-9}{4+3}=\frac{7}{7}=1$

Step 2. Apply the limit for c<4

On applying the limit, we get,

$\underset{x\to 4^{-}}{\lim }\frac{x^2-9}{x+3}=\underset{x\to 4^{-}}{\lim }\left(\frac{(x+3)(x-3)}{x+3}\right)$

$$\begin{gathered} =\underset{x\to 4^{-}}{\lim }(x-3) \\ =4-3 \\ =1 \end{gathered}$$

Step 3. Apply the limit for c>4

On applying the limit,

$$\begin{gathered} \underset{x\to 4^{+}}{\lim }(kx+5)=\underset{x\to 4^{+}}{\lim }kx+\underset{x\to 4^{+}}{\lim }5 \\ =k\underset{x\to 4^{+}}{\lim }x+\underset{x\to 4^{+}}{\lim }5 \\ =4k+5 \end{gathered}$$

Step 4. Value of k

We know, if a continuous limit exists then

$$\begin{gathered} \underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }f\left(x\right)=\underset{x\to 4}{\lim }f\left(x\right) \\ 1=4k+5 \\ -4=4k \\ -1=k \end{gathered}$$