We have a point at which \(\displaystyle f^{'}(x)=0\). We know from the second derivative test that if the second derivative is negative, the function has a maximum at that point. If the second derivative is positive, the function has a minimum at that point. If the second derivative is zero, the function has an inflection point at that point.

Plug in 0 into the second derivative to obtain 

\(\displaystyle f''(0)=0\)

So the point is an inflection point.

Notice that on the interval \(\displaystyle \left [ 0,1 \right ]\), the term \(\displaystyle \ x^{2}\) is always less than or equal to \(\displaystyle x\). So the function is largest at the points when \(\displaystyle x^2=x\). This occurs at \(\displaystyle x=0\) and \(\displaystyle x=1\).

Plugging in either 1 or 0 into the original function \(\displaystyle f(x)\) yields the correct answer of 0.

A cubic function will have at most one relative minimum and one relative maximum.  We can determine the zeros be factoring at \(\displaystyle x=-1,0,4\).  From then we only need to determine if the graph is positive or negative in-between the zeros. 

The graph is positive between \(\displaystyle -1\) and \(\displaystyle 0\) (plug in \(\displaystyle x=-0.5\)) and negative between 0 and 4 (plug in \(\displaystyle x=1\)).  This can also be seen from the graph.

The vertex form of a parabola is:

\(\displaystyle f(x)=a(x-h)^2+k\)

where \(\displaystyle {}(h,k)\) is the vertex of the parabola.

The function for this problem can be simplified into vetex form of a parabola: 

\(\displaystyle f(x)=(x-5)^2+1\) ,

with a vertex at \(\displaystyle (5,1)\).

Since the parabola is concave up, the minimum will be at the vertex of the parabola, which is at \(\displaystyle (5,1)\).

The average rate of change will be found by \(\displaystyle \frac{f(2)-f(0)}{2-0}\).

Here, \(\displaystyle f(2)=2e^2\), and \(\displaystyle f(0)=2e^0=2(1)=2\).

Now, we have \(\displaystyle \frac{2e^2-2}{2-0}=\frac{2e^2-2}{2}=e^2-1\).

We use the average rate of change formula, which gives us \(\displaystyle \frac{f(4)-f(0)}{4-0}\).

Now \(\displaystyle f(4)=-ln(4+1)=-ln5\), and \(\displaystyle f(0)=-ln(0+1)=-ln1=0\).

Therefore, the answer becomes \(\displaystyle \frac{-ln5-0}{4-0}=-\frac{ln5}{4}\).

We need to apply the formula for the average rate of change to our profit equation. Thus we find the average rate of change is \(\displaystyle \frac{P(4)-P(1)}{4-1}\).

Since \(\displaystyle P(4)=3(4)^2-4(4)+2=3(16)-16+2=48-16+2=34\), and \(\displaystyle P(1)=3(1)^2-4(1)+2=3(1)-4+2=3-4+2=1\), we find that the average rate of change is \(\displaystyle \frac{34-1}{4-1}=\frac{33}{3}=11\).

Since \(\displaystyle P(5)=2(5)^2-6(5)+25\), we see that this equals\(\displaystyle 2(25)-30+25=50-30+25=20+25=45\). Now let's examine \(\displaystyle P(4)\). \(\displaystyle P(4)=2(4)^2-6(4)+25=2(16)-24+25\) which simplifies to \(\displaystyle 33\).

Therefore the average rate of change formula gives us \(\displaystyle \frac{f(5)-f(4)}{5-4}=\frac{45-33}{5-4}=\frac{12}{1}=12\).

\(\displaystyle D(3)=15-9+3=6+3=9\)

\(\displaystyle D(2)=15-4+2=13\)

Thus the average rate of change formula yields \(\displaystyle \frac{9-13}{3-2}=\frac{-4}{1}=-4\).

This implies that the demand drops as the price increases.

\(\displaystyle A(3)=100e^{0.05*3}=100e^{0.15}\approx \$116.18\).

\(\displaystyle A(0)=100e^{0.05*0}=100e^{0}=\$100.00\).

Hence, the average rate of change formula gives us \(\displaystyle \frac{A(3)-A(0)}{3-0}=\frac{\$116.18-\$100.00}{3-0}=\frac{\$16.18}{3}\approx \$5.39\).