g(x) =(2x^2-3x-20)/(x-4), if x!=4 and =kx-15, if x=4 Evalu

UkusakazaL 2021-09-17 Answered
g(x)
\(\displaystyle=\frac{{{2}{x}^{{2}}-{3}{x}-{20}}}{{{x}-{4}}},{\quad\text{if}\quad}{x}\ne{4}\)
and
=kx-15, if x=4
Evaluate the constant k that makes the function continuous.

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Expert Answer

Arnold Odonnell
Answered 2021-09-18 Author has 8356 answers
For f(x) to be continuous at a point x=a:
\(\displaystyle{\underset{{{x}\to{a}^{+}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to{a}^{{-}}}}{{\lim}}}{f{{\left({x}\right)}}}={f{{\left({a}\right)}}}\)
It's given that g(x) is continuous at x=4, so
\(\displaystyle{\underset{{{x}\to{4}^{+}}}{{\lim}}}{g{{\left({x}\right)}}}={\underset{{{x}\to{4}^{{-}}}}{{\lim}}}{g{{\left({x}\right)}}}={g{{\left({4}\right)}}}\)
\(\displaystyle{\underset{{{x}\to{4}}}{{\lim}}}{g{{\left({x}\right)}}}={g{{\left({4}\right)}}}\)
\(\displaystyle{\underset{{{x}\to{4}}}{{\lim}}}\frac{{{2}{x}^{{2}}-{3}{x}-{20}}}{{{x}-{4}}}={k}{\left({4}\right)}-{15}\)
\(\displaystyle{\underset{{{x}\to{4}}}{{\lim}}}\frac{{{2}{x}^{{2}}-{8}{x}+{5}{x}-{20}}}{{{x}-{4}}}={4}{k}-{15}\)
\(\displaystyle{\underset{{{x}\to{4}}}{{\lim}}}\frac{{{2}{x}{\left({x}-{4}\right)}+{5}{\left({x}-{4}\right)}}}{{{x}-{4}}}={4}{k}-{15}\)
\(\displaystyle{\underset{{{x}\to{4}}}{{\lim}}}\frac{{{\left({2}{x}+{5}\right)}{\left({x}-{4}\right)}}}{{{x}-{4}}}={4}{k}-{15}\)
\(\displaystyle{\underset{{{x}\to{4}}}{{\lim}}}{\left({2}{x}+{5}\right)}={4}{k}-{15}\)
2(4)+5=4k-15
k=7
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