Let f(x)= 1/x and g(x)=x^2+5x.a. Find (f*g)(x)b. Find the domain and range of (f*g)(x)

we are given [tex]f(x)=\frac{1}{x}[/tex][tex]g(x)=x^2+5x[/tex](A)(f×g)(x)=f(x)*g(x)now, we can plug it [tex](fXg)(x)=\frac{1}{x} (x^2+5x)[/tex]we can simplify it [tex](fXg)(x)=\frac{1}{x} (x(x+5))[/tex][tex](fXg)(x)=x+5[/tex](B)Domain:Firstly, we will find domain  of f(x) , g(x) and (fxg)(x)and then we can find common domain Domain of f(x):[tex]f(x)=\frac{1}{x}[/tex]we know that f(x) is undefined at x=0so, domain will be [tex](-\infty,0)[/tex]∪[tex](0,\infty)[/tex]Domain of g(x):[tex]g(x)=x^2+5x[/tex]Since, it is polynomial so, it is defined for all real values of xnow, we can find common domain so, domain will be [tex](-\infty,0)[/tex]∪[tex](0,\infty)[/tex]..............AnswerRange:Firstly, we will find range of f(x) , g(x) and (fxg)(x)and then we can find common range Range of f(x):[tex]f(x)=\frac{1}{x}[/tex]we know that range is all possible values of y for which x is definedsince, horizontal asymptote will be at y=0so, range is [tex](-\infty,0)[/tex]∪[tex](0,\infty)[/tex]Range of g(x):[tex]g(x)=x^2+5x[/tex]Since, it is quadratic equation so, its range will be [tex][-6.25,\infty)[/tex]now, we can find common range so, range will be [tex](-6.25,0)[/tex]∪[tex](0,\infty)[/tex].............Answer