\(\displaystyle g(f(x))\) means \(\displaystyle f(x)\) gets plugged into \(\displaystyle g(x)\).

Thus \(\displaystyle g(f(x))= (3x^{2}+5) - 2 = 3x^{2}+3\).

Calculate \(\displaystyle g(2)\) and plug it into \(\displaystyle f(x)\).

\(\displaystyle g(2)= (2)^{2}-1=4-1=3\)

\(\displaystyle f(3)=(3)^{3}-2(3)^{2}+(3)=27-2(9)+(3)=27-18+3=12\)

\(\displaystyle \small f(g(3))\)

This expression is the same as saying "take the answer of \(\displaystyle \small g(3)\) and plug it into \(\displaystyle \small f(x)\)."

First, we need to find \(\displaystyle \small g(3)\). We do this by plugging \(\displaystyle \small 3\) in for \(\displaystyle \small x\) in \(\displaystyle \small g(x)\).

\(\displaystyle g(x)=x^2\)

\(\displaystyle g(3)=3^2=9\)

Now we take this answer and plug it into \(\displaystyle \small f(x)\).

\(\displaystyle f(g(3))=f(9)\)

We can find the value of \(\displaystyle \small f(9)\) by replacing \(\displaystyle \small x\) with \(\displaystyle \small 9\).

\(\displaystyle f(x)=6x-4\)

\(\displaystyle f(9)=6(9)-4=50\)

This is our final answer.

The function \(\displaystyle f(x+b)\) shifts a function f(x) \(\displaystyle b\) units to the left. Conversely, \(\displaystyle f(x-b)\) shifts a function f(x) \(\displaystyle b\) units to the right. In this question, we are translating the graph two units to the left.

To translate along the y-axis, we use the function \(\displaystyle f(x)+k\) or \(\displaystyle f(x)-k\).

We are asked to find \(\displaystyle f^{^{-1}} (x)\), which is the inverse of a function. 

In order to find the inverse, the first thing we want to do is replace f(x) with y. (This usually makes it easier to separate x from its function.).

\(\displaystyle y=5-(1+x)^3\)

Next, we will swap x and y.

\(\displaystyle x=5-(1+y)^3\)

Then, we will solve for y. The expression that we determine will be equal to \(\displaystyle f^{^{-1}} (x)\).

\(\displaystyle x=5-(1+y)^3\)

Subtract 5 from both sides.

\(\displaystyle x-5=-(1+y)^3\)

Multiply both sides by -1.

\(\displaystyle -1(x-5)=5-x=(1+y)^3\)

We need to raise both sides of the equation to the 1/3 power in order to remove the exponent on the right side. 

\(\displaystyle (5-x)^{\frac{1}{3}}=\left ((1+y)^{3} \right )^{1/3}\)

We will apply the general property of exponents which states that \(\displaystyle (a^b)^{c}=a^{b\cdot c}\).

\(\displaystyle (5-x)^{\frac{1}{3}}=\left ((1+y)^{3} \right )^{1/3}=(1+y)^{3\cdot \frac{1}{3}}=(1+y)^1=1+y\)

Laslty, we will subtract one from both sides.

\(\displaystyle (5-x)^{\frac{1}{3}}-1=y\)

The expression equal to y is equal to the inverse of the original function f(x). Thus, we can replace y with \(\displaystyle f^{-1}(x)\).

\(\displaystyle f^{-1}(x)=(5-x)^{\frac{1}{3}}-1\)

The answer is \(\displaystyle f^{-1}(x)=(5-x)^{\frac{1}{3}}-1\).

The inverse of \(\displaystyle y\) requires us to interchange \(\displaystyle x\) and \(\displaystyle y\) and then solve for \(\displaystyle y\).

\(\displaystyle y=4x^{2}\) 

\(\displaystyle x=4y^{2}\)

Then solve for \(\displaystyle y\):

\(\displaystyle y=\frac{\pm \sqrt{x}}{2}\)

To find the inverse of a function, exchange the \(\displaystyle x\) and \(\displaystyle y\) variables and then solve for \(\displaystyle y\).

\(\displaystyle x=y^{2}+1\)

\(\displaystyle x-1=y^{2}\)

\(\displaystyle \sqrt{x-1}=y=f^{-1}(x)\)

A horizontal line has infinitely many values for \(\displaystyle x\), but only one possible value for \(\displaystyle y\). Thus, it is always of the form \(\displaystyle y = c\), where \(\displaystyle c\) is a constant. Horizontal lines have a slope of \(\displaystyle 0\). The only equation of this form is \(\displaystyle y = 0\). 

A vertical line is one in which the \(\displaystyle y\) values can vary. Namely, there is only one possible value for \(\displaystyle x\), and \(\displaystyle y\) can be any number. Thus, by this description, the only vertical line listed is \(\displaystyle x = 3\). 

A line with a slope of zero will be horizontal. A horizontal line has only one possible value for \(\displaystyle x\), and \(\displaystyle y\) can be any value. 

Thus, the only given equation which fits this description is \(\displaystyle y = 10\).