The inflection points of a function are the points where the concavity changes, either from opening upwards to opening downwards or vice versa.  The inflection points occur at the x-values where the second derivative is either zero or undefined.  That means we need to find our second derivative.

We start by using the Power Rule to find the first derivative.

Taking the derivative once more gives the second derivative.

We then set this derivative equal to zero and solve.

This factors nicely.

Therefore our second derivative is zero when 

8 is the only one of these two amongst our choices and is therefore our answer.

To find inflection points, take the second derivative and set it equal to .

Since  is never , this function does not have any inflection points. Thus, the answer is none.

To solve, simply differentiate twice, find when the function is equal to 0, and then plug into the first equation.

For this particular function, use to power rule to differentiate.

The power rule states,

.

Also recall that the derivative of a constant is zero.

Applying the power rule and rule of a constant once to the function we can find the first derivative.

Thus,

From here apply the power rule and rule of a constant once more to find the second derivative of the function.

Now to solve for the inflection point, set the second derivative equal to zero.

From here plug this x value into the original function to find the y value of the inflection point.

Thus, our point is (0,1).

To solve, find when the 2nd derivative is equal to 0 and plug it into the first equation. We must use the power rule, as outlined below, to find each derivative.

Power rule:

Thus,

Since 2 is never equal to 0, there are no points of inflection.

To solve, you must set the second derivative equal to 0 and solve for x. To differentiate twice, use the power rule as outlined below:

Power Rule:

Therefore:

Remember, the derivative a constant is 0.

Now, set it equal to 0. Thus,

If we want to approximate the area under a curve using n=4, that means we will be using 4 rectangles. Because the problem asks us to approximate the area from x=0 to x=4, this means we will have a rectangle between x=0 and x=1, between x=1 and x=2, between x=2 and x=3, and between x=3 and x=4. We want to use a midpoint Riemann sum, so the height of each rectangle will be the value of the function at the midpoint of each interval:

Now that we know the height of each rectangle, all we have to do is find its area by multiplying the height by the width, which is just 1 for each rectangle. Then we add the area of all the rectangles to find our approximation for the area under the curve from x=0 to x=4:

A left hand approximation requires us to slice the region into four rectangles of equal width. Since the interval on which we are slicing is  and we want to create four rectangles of equal width, the width of each rectangle must be,

 .

The height of each rectangular slice is given by the function value at the left edge of each rectangle. Beginning at the leftmost edge of the first rectangle on the interval, these left endpoints exist at , but not at  as that is the right edge of the fourth and final rectangle on the interval.

Using the function given, , the first rectangle has height . The rest have height .

Multiply each by the width of each rectangle  to get the area of each rectangular slice. Then add the area of the slices together, and simplify to get the correct result.

The interval  divided into four sub-intervals gives rectangles with vertices of the bases at 

For the approximated area, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or , , , and 

Because each sub-interval has width , the approximated area using rectangles is

As such, the approximated area is

 units squared

The interval  divided into two sub-intervals gives two rectangles with vertices of the bases at 

For the approximated area, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or  and 

Because each sub-interval has width , the approximated area using rectangles is

As such, the approximated area is

 units squared

In order to approximate the area under a curve using rectangles, one must take the sum of the areas of discrete rectangles under the curve. Taking the height of each rectangle as the function evaluated at the left endpoint, we obtain the following rectangle areas:

The sum of the individual rectangles yields an overall area approximation of 100.