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.