# Determine the value of k that will make f(x) continuous everywhere.

Let

f(x)=\begin{cases}\frac{x^{2}-16}{x-4}~~ , if ~~x \neq 4\\kx-24,~~if ~~~ x=4 \end{cases}


## Step By Step Solution:

The given piecewise-defined function is:

f(x)=\begin{cases}\frac{x^{2}-16}{x-4}~~ , if ~~x \neq 4\\kx-24,~~if ~~~ x=4 \end{cases}


Notice that,

 f(x) =\frac{x^{2}-16}{x-4}

is not defined at x=4.But,

 \lim_{x \to 4}f(x)

is defined at x=4.Now,

f(x) =kx-24

is defined at x=4:

Using the Continuity of a Function,

\lim_{x \to a}f(x)=f(a)
\lim_{x \to 4}f(x)=f(4)


Simplify the expression for x=4:

f(x)=kx-24~~, at ~x=4 \\f(4)=4k-24

As we know,

\lim_{x \to 4}f(x)=f(4)


Substitute the

f(x)=\frac{x^{2}-16}{x-4} \\and \\f(4)=4k-24

into the above ,we get:

\lim_{x \to 4}{\left(\frac{x^{2}-16}{x-4}\right)}=4k-24


Evaluate the limit:

\lim_{x \to 4}{\left(\frac{x^{2}-16}{x-4}\right)}=4k-24
\implies\lim_{x \to 4}{\left(\frac{\cancel{(x-4)}(x+4)}{\cancel{x-4}}\right)}=4k-24~~~~~~~~~~~~{\left (a^2-b^2=(a-b)(a+b) \right)}
\implies\lim_{x \to 4}{\left(x+4\right)}=4k-24

Now , putting values as discontinuity is removed:

\implies8=4k-24
\implies4k=32
\therefore k=8