QUESTION 1
Given that:
[tex]f(x)=2x^2+3x-9[/tex],
[tex]g(x)=5x+11[/tex],
and
[tex]h(x)=-3x^2+1[/tex]
Then;
[tex]f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)[/tex]
[tex]f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1[/tex]
Group similar terms;
[tex]f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1[/tex]
Simplify;
[tex]f(x)-g(x)+h(x)=-x^2-2x-19[/tex]
QUESTION 2
Given that;
[tex]f(x)=4x-7[/tex].
[tex]g(x)=(x+1)^2[/tex]
and
[tex]s(x)=f(x)+g(x)[/tex]
Substitute the functions;
[tex]s(x)=4x-7+(x+1)^2[/tex]
Substitute x=3
[tex]s(3)=4(3)-7+(3+1)^2[/tex]
[tex]s(3)=12-7+(4)^2[/tex]
[tex]s(3)=5+16[/tex]
[tex]s(3)=21[/tex]
QUESTION 3
Given:
[tex]f(x)=3x+2[/tex]
[tex]g(x)=x^2-5x-1[/tex]
[tex]f(g(x))=f(x^2-5x-1)[/tex]
This implies that;
[tex]f(g(x))=3(x^2-5x-1)+2[/tex]
Expand the parenthesis;
[tex]f(g(x))=3x^2-15x-3+2[/tex]
[tex]f(g(x))=3x^2-15x-1[/tex]
QUESTION 4
The given function is;
[tex]f(x)=3(x-6)^2+1[/tex]
Let
[tex]y=3(x-6)^2+1[/tex]
[tex]\Rightarrow y-1=3(x-6)^2[/tex]
[tex]\Rightarrow \frac{y-1}{3}=(x-6)^2[/tex]
[tex]\Rightarrow \sqrt{\frac{y-1}{3}}=x-6[/tex]
[tex]\Rightarrow x=6+\sqrt{\frac{y-1}{3}}[/tex]
The range is:
[tex]\frac{y-1}{3}\ge0[/tex]
[tex]y-1\ge0[/tex]
[tex]y\ge1[/tex]
The interval notation is;
[tex][1,+\infty)[/tex]
