To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

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

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\)

Next, we must find the critical values, at which the first derivative is equal to zero:

\(\displaystyle \frac{x^2+4x}{(x+2)^2}=0\)

\(\displaystyle x(x+4)=0\)

\(\displaystyle c=0, -4\)

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

\(\displaystyle (-\infty, -4), (-4, -2), (-2, 0), (0, \infty)\)

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative, nor is it defined at \(\displaystyle x=-2\). 

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second and third intervals, the first derivative is negative, and on the fourth interval, the first derivative is positive. Thus, the function is increasing on \(\displaystyle (-\infty, -4)\cup(0, \infty)\). 

\(\displaystyle i(x)\) is the only function listed with a positive slope at \(\displaystyle x=5\). The sign of the slope at of a function at a point is determined by the value first derivative of the function at the corresponding point (i.e., \(\displaystyle f'(x)\), \(\displaystyle g'(x)\), \(\displaystyle h'(x)\), and \(\displaystyle i'(x)\)). 

\(\displaystyle i'(5)=-(5)^3+9(5)^2-3(5)-24\)

\(\displaystyle =-125+225-15-24=61\)

61 is positive, and so, \(\displaystyle i(x)\) has a positive slope at \(\displaystyle x=5\). The other derivative functions listed have non-positive outputs at \(\displaystyle x=5\).

The two curves intersect in between \(\displaystyle x=-3\) and \(\displaystyle x=4\), which can be found by solving the quadratic equation \(\displaystyle 12-x^2=-x\).

To solve for the area between curves, \(\displaystyle y=f(x)\) and \(\displaystyle y=g(x)\), we use the formula \(\displaystyle A=\int^b_a(f(x)-g(x))dx\)

For our problem:

\(\displaystyle A=\int^4_{-3}(12-x^2-(-x))dx\)

\(\displaystyle A=12x-\frac{x^3}{3}+\frac{x^2}{2}\) evaluated from \(\displaystyle -3\) to \(\displaystyle 4\) which yields \(\displaystyle \frac{343}{6}\).

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

\(\displaystyle \int_{2}^{7}\frac{1}{\sqrt{x+2}}dx=2\sqrt{x+2}\Big|_2^7=2\)

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by \(\displaystyle x=3\) and \(\displaystyle y=2\) in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve \(\displaystyle \frac{1}{9}x^{2}\) is \(\displaystyle \int_{0}^{3}\frac{1}{9}x^2=1\).

The area of the triangle above the curve \(\displaystyle y=x\) is 2. Therefore, the area bounded is \(\displaystyle 6-1-2=3\).

\(\displaystyle \int sin(x) dx = -cos(x) + C\)\(\displaystyle \int _0^\pi sin(x) dx = -cos(\pi) - -cos(0)= 1-(-1) = 2\)

The area of this region is given by the following integral:

\(\displaystyle Area = \int_{0}^{2} f(x) - g(x) = \int_{0}^{2} -x^{2} + 2x - (-sin(\frac{\pi x}{2}))\) or

\(\displaystyle Area = \int_{0}^{2} -x^{2} + 2x + sin(\frac{\pi x}{2})\)

Taking the antiderivative gives

\(\displaystyle -\frac{x^{3}}{3} + x^{2} -\frac{2}{\pi }cos(\frac{\pi x}{2})\), evaluated from \(\displaystyle x = 0\) to \(\displaystyle x = 2\).

\(\displaystyle F(2) = \frac{-8}{3} + 4 + \frac{2 }{\pi }\), and 

\(\displaystyle F(0) = -\frac{2}{\pi }\).

Thus, the area is given by:

\(\displaystyle F(2) - F(0) =- \frac{8}{3} + 4 + \frac{2}{\pi } - (-\frac{2}{\pi }) =\frac{4}{3} +\frac{4}{\pi }\)

By Rolle's Theorem, if \(\displaystyle f\) is continuous on \(\displaystyle [a,b]\) and differentiable on \(\displaystyle (a,b)\), and \(\displaystyle f(a)= f(b)\), then there must be \(\displaystyle c \in \left ( a,b \right )\) such that \(\displaystyle f'(c) = 0\). Nothing in the statement of this theorem addresses the location of the zeroes of the function itself. Therefore, the statement is false.

By the Mean Value Theorem (MVT), if a function \(\displaystyle f\) is continuous and differentiable on \(\displaystyle [a,b]\), then there exists at least one value \(\displaystyle c \in (a,b)\) such that \(\displaystyle f'(c) = \frac{f(b)-f(a)}{b-a}\). \(\displaystyle f\), a polynomial, is continuous and differentiable everywhere; setting \(\displaystyle a= -2, b= 2\), it follows from the MVT that there is \(\displaystyle c \in (-2, 2 )\) such that 

\(\displaystyle f'(c) = \frac{f(2)-f(-2)}{2-(-2)}\)

Evaluating \(\displaystyle f(-2)\) and \(\displaystyle f(2)\):

\(\displaystyle f(x) = x^{3} - 7x + 16\)

\(\displaystyle f(2) = 2^{3} - 7 (2) + 16 = 8 -14+16 = 10\)

\(\displaystyle f(-2) = (-2) ^{3} - 7 (-2) + 16 = -8 +14+16 = 22\)

The expression for \(\displaystyle f'(c)\) is equal to 

\(\displaystyle f'(c) = \frac{10-22}{2-(-2)} = \frac{-12}{4} = -3\),

the correct choice.

By Rolle's Theorem, if \(\displaystyle f\) is continuous on \(\displaystyle [a,b]\) and differentiable on \(\displaystyle (a,b)\), and \(\displaystyle f(a)= f(b)\), then there must be \(\displaystyle c \in \left ( a,b \right )\) such that \(\displaystyle f'(c) = 0\). 

\(\displaystyle f\) is given to be continuous. Also, if we set \(\displaystyle a=1, b=2\), we note that \(\displaystyle f(1) = f(2)=0\). This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be \(\displaystyle c \in (1,2)\) such that \(\displaystyle f'(c) = 0\).

Incidentally, it does follow from the given information that \(\displaystyle f(x)\) must have a zero on the interval \(\displaystyle (3,4)\), but this is due to the Intermediate Value Theorem, not Rolle's Theorem.