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

First, we find the second derivative:

The derivatives were found using the following rules:

, , ,

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

Note that the x values we find are limited by the interval given in the problem statement. 

Using the critical 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 positive, on the second interval, the second derivative is negative, and on the third interval, the second derivative is positive. Thus, two points of inflection exist where the second derivative changed in sign, . 

To find what code the computer generates, we must figure out the points of inflection for the function, which are the points at which the second derivative switches sign.

First, we find the second derivative:

The derivatives were found using the following rules:

, 

Next, we must find the x values - on the given interval - for which the second derivative is equal to zero:

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

Note that at the endpoints of each interval, 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. The sign of the second derivative changed twice, so points of inflection exist at .

The computer will generate the code for two points of inflection as .

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

First, we find the second derivative of the function:

and was found using the following rules:

, 

Next, we must the values of x for which the second derivative is equal to zero:

Using the critical 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 positive, on the second interval, it is negative, and on the third interval, it is positive. Thus, two points of inflection exist for the function, at  because the second derivative changed sign at each of these locations.

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

First, we find the second derivative of the function:

and was found using the following rules:

, 

Next, we must the values of x for which the second derivative is equal to zero:

Using the critical 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 the second derivative is positive. Thus, a point of inflection exists at  because the second derivative changed sign here. 

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

Taking the second derivative of the function, we get

The following rules were used:

, 

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

The values for which the second derivative is zero are imaginary numbers. The second derivative is always positive and never zero, thus never changing in sign; the function has no inflection points.

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.