We have a point at which . We know from the second derivative test that if the second derivative is negative, the function has a maximum at that point. If the second derivative is positive, the function has a minimum at that point. If the second derivative is zero, the function has an inflection point at that point.

Plug in 0 into the second derivative to obtain 

So the point is an inflection point.

Notice that on the interval , the term  is always less than or equal to . So the function is largest at the points when . This occurs at  and .

Plugging in either 1 or 0 into the original function  yields the correct answer of 0.

The domain is defined as the set of possible values for the x variable. In order to find the impossible values of x, we should:

a) Set the equation under the radical equal to zero and look for probable x values that make the expression inside the radical negative:

There is no real value for x that will fit this equation, because any real value square is a positive number i.e. cannot be a negative number.

b) Set the denominator of the fractional function equal to zero and look for probable x values:

Now we can solve the equation for x:

There is no real value for x that will fit this equation.

The radical is always positive and denominator is never equal to zero, so the f(x) is defined for all real values of x. That means the set of all real numbers is the domain of the f(x) and the correct answer is .

Alternative solution for the second part of the solution:

After figuring out that the expression under the radical is always positive (part a), we can solve the radical and therefore denominator for the least possible value (minimum value). Setting the x value equal to zero will give the minimum possible value for the denominator.

That means the denominator will always be a positive value greater than 1/2; thus it cannot be equal to zero by setting any real value for x. Therefore the set of all real numbers is the domain of the f(x).

The domain is defined as the set of all values of x for which the function is defined i.e. has a real result. The square root of a negative number isn't defined, so we should find the intervals where that occurs:

The square of any number is positive, so we can't eliminate any x-values yet.

If the denominator is zero, the expression will also be undefined.

Find the x-values which would make the denominator 0:

Therefore, the domain is .

The domain is defined as the set of all possible values of the independent variable . First,  must be greater than or equal to zero, because if , thenwill be undefined. In addition, the total expression under the radical, i.e.  must be greater than or equal to zero:

That means that the expression under the radical is always positive and therefore is defined. Hence, the domain of the function is .

A function's symmetry is related to its classification as even, odd, or neither.

Even functions obey the following rule:

Because of this, even functions are symmetric about the y-axis.

Odd functions obey the following rule:

Because of this, odd functions are symmetric about the origin.

If a function does not obey either rule, it is neither odd nor even. (A graph that is symmetric about the x-axis is not a function, because it does not pass the vertical line test.)

To test for symmetry, simply substitute into the original equation.

Thus, this equation is even and therefore symmetric about the y-axis.

To find the horizontal asymptote, compare the degrees of the top and bottom polynomials. In this case, the two degrees are the same (1), which means that the equation of the horizontal asymptote is equal to the ratio of the leading coefficients (top : bottom). Since the numerator's leading coefficient is 1, and the denominator's leading coefficient is 2, the equation of the horizontal asymptote is .

To find the vertical asymptote, set the denominator equal to zero to find when the entire function is undefined:

This question is asking for the equation's slant asymptote. To find the slant asymptote, divide the numerator by the denominator. Long division gives us the following:

However, because we are considering  as it approaches infinity, the effect that the last term has on the overall linear equation quickly becomes negligible (tends to zero). Thus, it can be ignored.