Answer
$f(x)=x^{2}+6$
$g(x)=x^{2}+x-1$
Work Step by Step
First problem:
Squaring g(x), ... $[g(x)]^{2}=4x^{2}+4x+1,$
so we need to add 6 to this and we have h(x).
Our function f will "square the number and add 6", that is $f(x)=x^{2}+6$
In the second problem, we search for g(x) such that $f(g(x))=h(x),$
$3[g(x)]+5=3x^{2}+3x+2$
Equality of polynomials $\Rightarrow$ equate the coefficients.
g is quadratic, and the leading coefficient is 1.
g(x)$=x^{2}+bx+c$
From $3c+5 $(off the left side) being equal to 2 (on the right side),
we find $c=-1,$ so,
$g(x)=x^{2}+bx-1$
Since 3b must equal 3, it follows that b=1.
So, $g(x)=x^{2}+x-1$
Check:
$f(g(x))=3(x^{2}+x-1)+5=3x^{2}+3x+2=h(x)$
