To determine the function's points of inflection, we must determine the points at which the second derivative of the function changes in sign.

First, we must find the second derivative of the function:

The following rules were used:

, 

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

Using this value, we now create intervals over which to evaluate the sign of the second derivative:

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, but on the second interval, the second derivative is positive. The second derivative changes sign at , so a point of inflection exists here.

Points of inflection occur where the second derivative switches sign, which occurs when the second derivative is 0. Solving for y'',

Setting the second derivative to 0,

Before assuming this is a point of inflection, however, we need to make sure the second derivative actually changes sign. Plugging in a value to the left of 1 and a value to the right of 1 reveals

 and .

Note that the second derivative didn't change sign! More conceptually, this is because the (x-1) term is squared, and a squared number is never negative. Thus, while the second derivative hit 0, it immediately went back upwards, meaning it did not change from concave up to concave down. Thus, the function has no points of inflection.

To determine the x-values at which the function has points of inflection, we must determine the points at which the second derivative of the function changes sign.

First, we must find the second derivative of the function,

which was found using the following rules:

, 

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

We stopped finding the x values where they are bounded by the interval given in the problem statement.

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

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, on the second interval, the second derivative is positive, and on the third interval, the second derivative is negative. Thus, points of inflection exist at . 

To determine the x-values at which the function has points of inflection, we must determine the points at which the second derivative of the function changes sign.

First, we must find the second derivative of the function,

which was found using the following rules:

, 

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

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

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, and on the second interval, the second derivative is positive. Thus, a point of inflection occurs at , and because this the only inflection point - and it occurs at a negative x value - the computer reports .

For this question we want to find the points at which the second derivative of the function changes sign, i.e. . These are points at which the slope of the first derivative changes signs.

Notice that there are two x-values at which the second derivative equals zero.

We are interested only in the point where the slope changes from decreasing to increasing, however. This means that the answer is at .

x-values less than  (and greater than  ) yield negative y-values, and x-values greater than  yield positive y-values. In other words,  is the only point at which the second derivative changes from negative to positive, which represents the slope (first derivative) changing from decreasing to increasing.

To find the coordinates of the local extrema of a function, we need to find the critical points of its first derivative.

Since is a polynomial, we can find its derivative term by term. The first 3 terms can be differentiated using the power rule, , and the constant multiple rule, .

The last term is a constant, and its derivative is zero.

Applying these rules, we find the first derivative:

Now we need to find the critical points. To do this, we set the first derivative equal to zero and solve. Factoring is the best method in this problem.

Now that we have the critical points, we need to determine for each one, whether it is a maximum, minimum, or neither. We use the first derivative line test to determine this.

For , we will test the interval before it, , and the interval after it, , and find whether they are increasing or decreasing.

For the interval , we will test and find whether is positive or negative.

Since is positive, the original function is increasing before the critical point, .

Now we will test the interval after .

For the interval,  , we will test and find whether is positive or negative.

Since is negative, the original function is decreasing in the interval following .

Since the function is increasing before , and decreasing afterward, we can conclude that a maximum occurs at .

Now we find the value of this maximum, .

Thus  is a local maximum.

Now we will determine whether a maximum or minimum occurs at .

We know that is decreasing before , but we still need to determine what happens afterward.

For the interval , we will test , and find whether is positive or negative.

Since is positive, the original function is increasing following .

Since is decreasing before, and increasing after , we can conclude that a minimum occurs at .

Now we need to find the value of this minimum, .

Thus is a local minimum.

So our answer is:

is a local maximum.

is a local minimum.

Concavity refers to the "curving" of the function. While the first derivative describes when the function is increasing or decreasing (instantaneous rate of change of ), the second derivative describes concavity, the instantaneous rate of change of .

The first derivative is like velocity, (moving forward or backward),  while the second derivative is like acceleration (speeding up or slowing down).

Since we need to find the intervals of concavity, we will find the second derivative and work with it.

First we must find the first derivative using the power rule and constant multiple rule for each term of .

This gives:

Now to find the second derivative, we take the derivative of .

The same derivative rules apply:

Now that we have the second derivative, , we must find its critical points. We do this by setting , and solving.

The best method for this case is factoring.

The greatest common factor is ,

Inside the parentheses is a quadratic expression that can be factored like so:

Setting each factor equal to zero we find the following:

Now we know the critical points for the second derivative. To find the intervals of concavity, we test a point in each interval around these critical points and find whether the second derivative is positive or negative in that interval.

For the interval , we can test .

Using the factored form of the second derivative is easier than using the polynomial form, since the arithmetic involves fewer large numbers.

Since is negative, the original function is Concave Down in the interval .

Now for the interval , we can test .

We will combine the fractions inside the parentheses by getting the common denominator.

Multiplying the fractions gives:

Since is positive, the original function is Concave Up on the interval .

For the interval , we can test .

Since is negative, the original function is Concave Down on the interval .

Finally, for the interval , we can test .

Since is positive, the original function is Concave Up on the interval .

Summarizing the results, the intervals of concavity are:

Concave Down: 

Concave Up:

To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

Remember that anything to the zero power is one.

Plug in our given value.

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use  as our expression.

For this problem, we're going to say that  since, as stated before, anything to the zero power is one.

Notice that  as anything times zero is zero.

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

To answer this question, one first needs to find  and then find the critical points of the function (i.e. where .  Finally, one would need to determine the sign of  for the intervals between the critical points.  

For the given function:

 .

Therefore,  when  and .  So, the intervals to consider are:

To determine the sign of , pick any number for the given interval and evaluate  at that number.

Therefore,  is increasing on the intervals  and  since  is greater than zero on these intervals.

On the domain , we know that the derivative begins positive, and because the concavity is positive, we know the derivative is increasing. Thus, the derivative stays positive for this entire interval, and the function increases from 2 to 4. Thus,  must be greater than .

In the case of the interval , we know that the derivative is increasing, but it starts out negative. Thus, perhaps the derivative only increased from -1 to -0.5 in this interval, and the function would have decreased the entire time. In this case,  would be less than , so we can't really say anything about these values.

For the remaining two, there's not any clear way to relate the functions at  and . While we know  needs to be bigger than 1, we don't know by how much. Similarly, while we know  needs to be bigger than -1, we don't know by how much. Thus, it's completely possible that  and .

As for  and , we know even less. In between the two intervals, our function could have shot up a million, or shot down by the same amount. Thus, there's no safe comparison we can make between these two values.