Answer:
1) [tex](f o g)(x)=\frac{4}{x+6}[/tex]
2) [tex](f*g)(x)=\frac{2x^2+9x^{\frac{3}{2}}+9}{2x^2}[/tex]
3) [tex](f+g)(x)=5x^3-2x^2-5x+8[/tex]
Step-by-step explanation:
1) Given : [tex]f(x)=\frac{2}{x+3}[/tex] and [tex]g(x)=\frac{1}{2}x[/tex]
To find : [tex](f o g)(x)[/tex]
Solution :
We know that,
[tex](f o g)(x)=f(g(x))[/tex] i.e g(x) within f(x),
[tex]f(g(x))=\frac{2}{g(x)+3}[/tex]
[tex]f(g(x))=\frac{2}{\frac{1}{2}x+3}[/tex]
[tex]f(g(x))=\frac{2}{\frac{x+6}{2}}[/tex]
[tex]f(g(x))=\frac{4}{x+6}[/tex]
Therefore, [tex](f o g)(x)=\frac{4}{x+6}[/tex]
2) Given : [tex]f(x)=\sqrt x+\frac{3}{x}[/tex] and [tex]g(x)=\sqrt x+\frac{3}{2x}[/tex]
To find : [tex](f*g)(x)[/tex]
Solution :
We know that,
[tex](f*g)(x)=f(x) \times g(x)[/tex]
[tex]f(x) \times g(x)=(\sqrt x+\frac{3}{x})\times (\sqrt x+\frac{3}{2x})[/tex]
[tex]f(x) \times g(x)=x+\frac{3\sqrt x}{2x}+\frac{3\sqrt x}{x}+\frac{9}{2x^2}[/tex]
[tex]f(x) \times g(x)=\frac{2x^2+3x\sqrt x+6x\sqrt x+9}{2x^2}[/tex]
[tex]f(x) \times g(x)=\frac{2x^2+9x^{\frac{3}{2}}+9}{2x^2}[/tex]
Therefore, [tex](f*g)(x)=\frac{2x^2+9x^{\frac{3}{2}}+9}{2x^2}[/tex]
3) Given : [tex]f(x)=x^3-2x^2+1[/tex] and [tex]g(x)=4x^3-5x+7[/tex]
To find : [tex](f+g)(x)[/tex]
Solution :
We know that,
[tex](f+g)(x)=f(x)+g(x)[/tex]
[tex]f(x)+g(x)=x^3-2x^2+1+4x^3-5x+7[/tex]
[tex]f(x)+g(x)=5x^3-2x^2-5x+8[/tex]
Therefore, [tex](f+g)(x)=5x^3-2x^2-5x+8[/tex]
