Respuesta :
Answer:
a.[tex](f+g)(-4)=-10[/tex]
b.[tex](f-g)(0)=\frac{1}{\sqrt{5} } -1[/tex]
c.[tex](fg)(-1)=-1[/tex]
d.[tex](\frac{f}{g})(1)=\frac{1}{4\sqrt{6} }[/tex]
Step-by-step explanation:
Given that:
[tex]f(x)=\frac{1}{\sqrt{x+5} }[/tex]
[tex]g(x)=3x+1[/tex]
The sum, rest, multiplication or division of functions are calculated as following:
[tex](f+g)(x)=f(x)+g(x)\\(f-g)(x)=f(x)-g(x)\\(fg)(x)=f(x)g(x)\\(\frac{f}{g} )(x)=\frac{f(x)}{g(x)}[/tex]
For values of x where f(x) and g(x) are defined and in the case of [tex](\frac{f}{g} )(x)[/tex] for values of g(x) different from zero.
Taking into account that f(x) and g(x) are defined for values of x equals to -4, 0, 1 and -1 and g(1) is different from zero, we get:
[tex](f+g)(x)=f(x)+g(x)\\(f+g)(x)=\frac{1}{\sqrt{x+5} } +3x+1\\(f+g)(-1)=\frac{1}{\sqrt{-4+5} } +3(-4)+1\\(f+g)(-1)=-10[/tex]
[tex](f-g)(x)=f(x)-g(x)\\(f-g)(x)=\frac{1}{\sqrt{x+5} } -(3x+1)\\(f-g)(0)=\frac{1}{\sqrt{0+5} } -(3(0)+1)\\(f-g)(0)=\frac{1}{\sqrt{5} } -1[/tex]
[tex](fg)(x)=f(x)g(x)\\(fg)(x)=\frac{1}{\sqrt{x+5} } (3x+1)\\(fg)(-1)=(\frac{1}{\sqrt{-1+5} } )(3(-1)+1)\\(fg)(-1)=-1[/tex]
[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)} \\(\frac{f}{g})(x)=\frac{1}{(\sqrt{x+5})(3x+1)} \\(\frac{f}{g})(1)=\frac{1}{(\sqrt{1+5})(3(1)+1)}\\(\frac{f}{g})(1)=\frac{1}{4\sqrt{6} }[/tex]
