Answer:
g(x)<j(x)<k(x)<f(x)<m(x)<h(x)
Step-by-step explanation:
1.[tex]f(x)=\frac{x^2+x-20}{x^2+4}[/tex]
The denominator of f is defined for all real values of x
Therefore, the function is continuous on the set of real numbers
[tex]\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345[/tex]
3.[tex]h(x)=\frac{3x-5}{x^2-5x+7}[/tex]
[tex]x^2-5x+7=0[/tex]
It cannot be factorize .
Therefore, it has no real values for which it is not defined .
Hence, function h is defined for all real values.
[tex]\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43[/tex]
2.[tex]g(x)=\frac{x-17}{x^2+75}[/tex]
The denominator of g is defined for all real values of x.
Therefore, the function g is continuous on the set of real numbers
[tex]\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12[/tex]
4.[tex]i(x)=\frac{x^2-9}{x-9}[/tex]
x-9=0
x=9
The function i is not defined for x=9
Therefore, the function i is not continuous on the set of real numbers.
5.[tex]j(x)=\frac{4x^2-7x-65}{x^2+10}[/tex]
The denominator of j is defined for all real values of x.
Therefore, the function j is continuous on the set of real numbers.
[tex]\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0[/tex]
6.[tex]k(x)=\frac{x+1}{x^2+x+29}[/tex]
[tex]x^2+x+29=0[/tex]
It cannot be factorize .
Therefore, it has no real values for which it is not defined .
Hence, function k is defined for all real values.
[tex]\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102[/tex]
7.[tex]l(x)=\frac{5x-1}{x^2-9x+8}[/tex]
[tex]x^2-9x+8=0[/tex]
[tex]x^2-8x-x+8=0[/tex]
[tex]x(x-8)-1(x-8)=0[/tex]
[tex](x-8)(x-1)=0[/tex]
[tex]x=8,1[/tex]
The function is not defined for x=8 and x=1
Hence, function l is not defined for all real values.
8.[tex]m(x)=\frac{x^2+5x-24}{x^2+11}[/tex]
The denominator of m is defined for all real values of x.
Therefore, the function m is continuous on the set of real numbers.
[tex]\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722[/tex]
g(x)<j(x)<k(x)<f(x)<m(x)<h(x)
