# Find the value of k for which the function is continuous everywhere:

## Find the value of k for which the function is continuous everywhere.

Step-by-step solution-:

The given function is:

The given piecewise-defined function consists of two sub-functions that are one logarithmic

f(x)=\log_{3}{(x+1)}+\log_{3}{(x+3)},x\geq0

another polynomial function is:

f(x)=x^{2}-k , x<0

Calculate the Left-hand side limit for f(x)=x^2-k:

\lim_{x\to{0}^{-}}{(x^{2}-k)}

After simplifying the above limit:

We get,

\lim_{x\to{0}^{-}}{(x^{2}-k)}=(0^2-k)=-k

Calculate the Right-hand side limit for f(x)=\log_{3}{(x+1)}+\log_{3}{(x+3)}:

\lim_{x\to{0}^{+}}{\left(\log_{3}{(x+1)}+\log_{3}{(x+3)}\right)}

After simplifying the above limit:

We get,

\lim_{x\to{0}^{+}}{\left(\log_{3}{(x+1)}+\log_{3}{(x+3)}\right)}=\log_{3}{(0+1)}+\log_{3}{(0+3)}

After simplifying

\log_{3}{(0+1)}+\log_{3}{(0+3)}=\log_{3}{(1)}+\log_{3}{(3)}

We know that log(1) of bae 3=0, log(3) of base 3=1:

\log_{3}{(1)}+\log_{3}{(3)}=0+\log_{3}{(3)}=\log_{3}{(3)}=1

Since the consists of two sub-functions that are one logarithmic and another polynomial function which are continuous on the domain, When the Left-hand limit and right-hand limit are equal:

\lim_{x\to{0}^{-}}{(x^{2}-k)}=\lim_{x\to{0}^{+}}{\left(\log_{3}{(x+1)}+\log_{3}{(x+3)}\right)}

Now,

Find the value of k for which the function is continuous.

Substitute the value of the left-hand limit and right-hand limit into the above:

-k=1

After simplifying we get:

k=-1

The required value of k is -1: