
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:
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