The derivative of the function  is .  There is no local maximum for this plot.  Nowhere in the plot of the derivative is there a point where the derivative crosses the x-axis going from positive to negative.

We may want to answer true because we associate concavity with increasing functions, but functions that are decreasing can also be concave up.  The rule of thumb is, if the derivative of a function is increasing on an interval, then the function is concave up.  Here, we are only given the information that the derivative is concave up, but no insight into whether or not it is increasing or decreasing.  So this statement is false.

Remember that critical points of a derivative give us insight into maximum, minimums, and points of inflection.  If the derivative crosses the x-axis going from positive to negative then this tells us that the function has a local maximum at this point.  If the derivative crosses the x-axis going from negative to positive, then this tells us that the function has a local minimum at that point.  Remember the derivative tells us about the rate of change.  If the slope of our function changes from negative to positive, then there must be a small trench that is a minimum.  If the slope of our function changes from positive to negative, then there must be a small hill that is a maximum.

This is true.  For all points of the derivative that are above the x-axis, this is telling us the slope of our function is positive and therefore increasing.  For all points of the derivative below the x-axis, this is telling us the slope of our function is negative and therefore decreasing.

From the description of the derivative we know that a local maximum is at .  Since the derivative is crossing the x-axis going from positive to negative at this point, we assume that the slope of the function  is going from positive to negative creating a local maximum.  At  we are give that the derivative crosses the x-axis going from negative to positive.  This means that our function  is going from decreasing to increasing creating a local minimum.  The graph above meets these criteria.

We know that our function will be concave up when the derivative is increasing and concave down when the derivative is decreasing.  The graph of the derivative is increasing on the interval  and so  will be concave up on this interval.  The graph of the derivative is decreasing on the interval  and so  will be concave down on this interval.

Just because a derivative is increasing does not mean that the function  will be positive.  We know that if the derivative is increasing, then the function must be concave up.  We are given no insight as to where the function starts and no insight as to whether the function is positive or negative.

At a relative minimum  of the graph , it will hold that  and . 

First, find . Using the sum rule,

Differentiate the individual terms using the Constant Multiple and Power Rules:

Set this equal to 0:

Either:

, in which case, ; this equation has no real solutions.

 has two real solutions,  and . 

Now take the second derivative, again using the sum rule:

Differentiate the individual terms using the Constant Multiple and Power Rules:

Substitute  for :

Therefore,  has a relative minimum at .

Now. substitute  for :

Therefore,  has a relative maximum at .

Because our  is constant, the left Riemann sum will be

To use left Riemann sums, we need to use the following formula:

.

where  is the number of subintervals, (4 in our problem),

 is the "counter" that denotes which subinterval we are working with,(4 subintervals mean that  will be 1, 2, 3, and then 4)

 is the function value when you plug in the "i-th" x value, (i-th in this case will be 1-st, 2-nd, 3-rd, and 4-th)

, is the width of each subinterval, which we will determine shortly.

and  means add all  versions together (for us that means add up 4 versions).

This fancy equation approximates using boxes. We can rewrite this fancy equation by writing , 4 times; 1 time each for , , , and . This gives us

Think of  as the base of each box, and  as the height of the 1st box.

This is basically , 4 times, and then added together.

Now we need to determine what  and  are.

To find  we find the total length between the beginning and ending x values, which are given in the problem as  and . We then split this total length into 4 pieces, since we are told to use 4 subintervals.

In short, 

Now we need to find the x values that are the left endpoints of each of the 4 subintervals. Left endpoints because we are doing Left Riemann sums.

The left most x value happens to be the smaller of the overall endpoints given in the question. In other words, since we only care about the area from  to , we'll just use the smaller one, , for our first .

Now we know that .

To find the next endpoint, , just increase the first x by the length of the subinterval, which is . In other words

Add the  again to get 

And repeat to find 

Now that we have all the pieces, we can plug them in.

plug each value into  and then simplify.

This is the final answer, which is an approximation of the area under the function.