Whenever there is a function, all you need to do is plug in the \(\displaystyle x-\)value into the function. Since this is multi-function, whichever answer we get for the inside function, we plug it into the outer function.

\(\displaystyle f(x)=3x\) ; \(\displaystyle g(x)=x^3+1\)

\(\displaystyle f(g(3))=f(3^3+1)=f(28)\)

\(\displaystyle f(28)=3(28)=84\)

Whenever there is a function, all you need to do is plug in the \(\displaystyle x-\)value into the function. Since this is multi-function, whichever answer we get for the inside function, we plug it into the outer function.

\(\displaystyle f(x)=x^2-2\) and \(\displaystyle g(x)=x+4\)

\(\displaystyle g(f(x+1))=g((x+1)^2-2)=g(x^2+x+x+1-2)=g(x^2+2x-1)\)

\(\displaystyle g(x^2+2x-1)=x^2+2x-1+4=x^2+2x+3\)

We know \(\displaystyle f(x)=3x+2\) and \(\displaystyle f(x)=14\). Just set them equal to each other.

\(\displaystyle 3x+2=14\) Subtract \(\displaystyle 2\) on both sides.

\(\displaystyle 3x=12\) Divide \(\displaystyle 3\) on both sides.

\(\displaystyle x=4\)

We know \(\displaystyle f(x)=x^2+17\) and \(\displaystyle f(x)=66\). Just set them equal to each other.

\(\displaystyle x^2+17=66\) Subtract \(\displaystyle 17\) on both sides.

\(\displaystyle x^2=49\) Take square root on both sides and account for also negative answers.

\(\displaystyle x=\pm7\)

We know that \(\displaystyle f(x)=\left | x+1\right |\) and \(\displaystyle f(x)=12\). Just set them equal to each other.

\(\displaystyle \left | x+1\right |=12\) Remember to account for negative values.

\(\displaystyle x+1=12\) Subtract \(\displaystyle 1\) on both sides.

\(\displaystyle x=11\)

\(\displaystyle x+1=-12\) Subtract \(\displaystyle 1\) on both sides.

\(\displaystyle x=-13\)

We know \(\displaystyle f(g(x))=30\) so we need to apply substitutions to solve for \(\displaystyle x\).

\(\displaystyle f(g(x))=f(x^2)=x^2+5=30\) Subtract \(\displaystyle 5\) on both sides.

\(\displaystyle x^2=25\) Take square root on both sides and account for negative values.

\(\displaystyle x=\pm5\)

\(\displaystyle f(x)=x-1, g(x)=x^2+x\)

We know \(\displaystyle f(g(x))=5\) so let's make the substitution.

\(\displaystyle f(g(x))=f(x^2+x)=x^2+x-1=5\) This is a quadratic so subtract \(\displaystyle 5\) on both sides.

\(\displaystyle x^2+x-6=0\) Factor.

\(\displaystyle (x+3)(x-2)=0\) Solve individually.

\(\displaystyle x=-3, 2\)

By definition,

\(\displaystyle (f \circ g) (x) = f [g(x)]\), 

so

\(\displaystyle f [g(x)] = 14x + 4\)

If

\(\displaystyle f(x) = 7x - 15\),

it follows that 

\(\displaystyle f [g(x)] =7g(x) - 15\),

and, substituting, 

\(\displaystyle 7g(x) - 15 = 14x + 4\)

Solving for \(\displaystyle g(x)\) by isolating this expression:

\(\displaystyle 7g(x) - 15 + 15 = 14x + 4 + 15\)

\(\displaystyle 7g(x) = 14x + 19\)

\(\displaystyle 7g(x) \div 7 = (14x + 19) \div 7\)

\(\displaystyle g(x) = \frac{14x + 19}{7}\)

\(\displaystyle g(x) = 2x + \frac{ 19}{7}\).

By definition,

\(\displaystyle (f \circ g) (x) = f [g(x)]\), 

so

\(\displaystyle f [g(x)] = 6x\)

If

\(\displaystyle f(x) = x^{2} + 4\),

it follows that 

\(\displaystyle f [g(x)] = [g(x)]^{2} + 4\),

and, substituting, 

\(\displaystyle [g(x)]^{2} + 4 = 6x\)

Solving for \(\displaystyle g(x)\) by isolating this expression:

\(\displaystyle [g(x)]^{2} + 4 = 6x\)

\(\displaystyle [g(x)]^{2} = 6x - 4\)

Taking the square root of both sides:

\(\displaystyle g(x) = \pm \sqrt{ 6x - 4}\)

Either \(\displaystyle g(x) =- \sqrt{ 6x - 4}\), which is not among the given choices, or \(\displaystyle g(x) = \sqrt{ 6x - 4}\), which is.

By definition,

\(\displaystyle (f \circ g) (x) = f [g(x)]\), 

so

\(\displaystyle f [g(x)] = 9x\)

If

\(\displaystyle f(x) = \sqrt{3x+ 21}\),

it follows that 

\(\displaystyle f [g(x)] = \sqrt{3 [g(x)]+ 21}\),

and, substituting, 

\(\displaystyle \sqrt{3 [g(x)]+ 21 } = 9x\)

Solving for \(\displaystyle g(x)\) by isolating this expression:

\(\displaystyle \left (\sqrt{3 [g(x)]+ 21} \right ) ^{2}=\left ( 9x \right )^{2}\)

Applying the Power of a Product Rule:

\(\displaystyle 3 [g(x)]+ 21= 9^{2} \cdot x^{2}\)

\(\displaystyle 3 [g(x)]+ 21= 81 x^{2}\)

\(\displaystyle 3 [g(x)]+ 21- 21 = 81 x^{2} - 21\)

\(\displaystyle 3 [g(x)] = 81 x^{2} - 21\)

\(\displaystyle \frac{3 [g(x)] }{3} = \frac{ 81 x^{2} - 21 }{3}\)

\(\displaystyle g(x) = 27 x^{2} - 7\)