We will find the points of inflection by setting our second derivative to zero.

Take the derivative again,

then set this equal to zero and reverse foil,

These are our points of inflection.

To find the inflection points of , we need to find  (which lucky for us, is already given!) set it equal to , and solve for .

. Start

. Divide by . We can do this, because  is never equal to .

On the unit circle, the values  cause , but only  is inside our interval . so  is the only value to consider here.

To prove that  is actually part of a point of inflection, we have to test an  value on the left and the right of , and substitute them into  and test their signs.

.

Hence , is the  coordinate of an inflection point of  on the interval .

Inflection points of a function occur when the second derivative equals zero. Therefore, we simply need to take two derivatives of our function, and solve.

Therefore, the two values that makes this function go to zero are 

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:

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 negative, while on the second interval, the second derivative is positive. The second derivative changed sign , so there exists the point of inflection for the function. 

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.