Answer:
[tex]\begin{gathered} (a)(f-g)(x)=2x^2+x-8 \\ (b)(f\cdot g)(x)=4x^3+20x^2+19x-7 \\ (c)f(x+h)-f(x)=4xh+2h^2+3h \\ (d)f(g(x))=8x^2+62x+118 \\ (e)g^{-1}(x)=\frac{x-7}{2} \end{gathered}[/tex]Explanation:
Give f(x) and g(x) below:
[tex]\begin{gathered} f(x)=2x^2+3x-1 \\ g(x)=2x+7 \end{gathered}[/tex](a) (f-g)(x)
[tex]\begin{gathered} (f-g)(x)=f(x)-g(x) \\ =[2x^2+3x-1]-[2x+7] \\ =2x^2+3x-1-2x-7 \\ =2x^2+3x-2x-1-7 \\ \implies(f-g)(x)=2x^2+x-8 \end{gathered}[/tex](b) (fg)(x)
[tex]\begin{gathered} (f\cdot g)(x)=f(x)g(x) \\ =[2x^2+3x-1][2x+7] \\ =2x^2[2x+7]+3x[2x+7]-1[2x+7] \\ =4x^3+14x^2+6x^2+21x-2x-7 \\ \implies(f\cdot g)(x)=4x^3+20x^2+19x-7 \end{gathered}[/tex](c) f(x+h)-f(x)
[tex]\begin{gathered} f(x)=2x^{2}+3x-1 \\ f(x)=2(x+h)^2+3(x+h)-1 \end{gathered}[/tex]Therefore:
[tex]f(x+h)-f(x)=[2(x+h)^2+3(x+h)-1]-[2x^2+3x-1][/tex]We then simplify below:
[tex]\begin{gathered} =2(x^2+2xh+h^2)+3x+3h-1-2x^2-3x+1 \\ =2x^2+4xh+2h^2+3x+3h-1-2x^2-3x+1 \\ =2x^2-2x^2+4xh+2h^2+3x-3x+3h-1+1 \\ =4xh+2h^2+3h \end{gathered}[/tex]Therefore:
[tex]f(x+h)-f(x)=4xh+2h^2+3h[/tex](d) f(g(x))
[tex]\begin{gathered} f(x)=2x^2+3x-1 \\ f[g(x)]=2[g(x)]^2+3[g(x)]-1 \\ \implies f[g(x)]=2[2x+7]^2+3[2x+7]-1 \end{gathered}[/tex]We simplify the resulting expression below:
[tex]\begin{gathered} f[g(x)]=2[(2x+7)(2x+7)]^+6x+21-1 \\ =2[4x^2+14x+14x+49]+6x+20 \\ =2[4x^2+28x+49]+6x+20 \\ =8x^2+56x+98+6x+20 \\ =8x^2+56x+6x+98+20 \\ \implies f(g(x))=8x^2+62x+118 \end{gathered}[/tex](e)g^(-1)(x)
[tex]\begin{gathered} g(x)=2x+7 \\ Swap\text{ x and g\lparen x\rparen} \\ x=2g^{-1}(x)+7 \\ \text{ Solve the equation for }g^{-1}(x) \\ 2g^{-1}(x)=x-7 \\ \implies g^{-1}(x)=\frac{x-7}{2} \end{gathered}[/tex]
