To find the slope of a line tangent to a parabola at a specific point, find the derivative of the parabola's equation, then substitute the -coordinate of the specific point in the new equation.

In this case, it helps to expand the equation before taking the derivative:

Now take the derivative of the expanded equation:

Since the -intercept is the point where the -coordinate is , substitute  into the equation for .

To find potential minima of the function, take the first derivative of  using the power rule.

Set the derivative to 0:

We solve for  to obtain  and then plug in 0.5 into the original function to obtain the answer of

We can double check that  is indeed a minimum by using the second derivative test

which means the function is concave up, so that the point we found is a minimum.

To find the maximum, we need to look at the first derivative. 

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat  as  since anything to the zero power is one.

Notice that  since anything times zero is zero.

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Notice that  changes from positive to negative when .

We can find that root using the quadratic equation: 

Since we're looking for a negative value, we will subtract.

Therefore, the maximum is at .

To find the maximum, we must find where the graph shifts from increasing to decreasing. To find out the rate at which the graph shifts from increasing to decreasing, we look at the second derivative and see when the value changes from positive to negative.

That is to say, we will look at the second derivative and see where (if at all) the graph crosses the x-axis and is moving from a positive y-value to a negative y-value.

Now we must find the second derivative. Unfortunately, the derivatives of trig functions must be memorized. The first derivative is:

To find the local minimum of , we need to look at the first derivative. 

To find the first derivative, we can use the power rule. The power rule states that we multiply each variable by its current exponent and then lower that exponent by one. 

Simplify.

Anything to the zero power is one, so .

Therefore, .

At the minimum, our  graph will cross the -axis. Therefore, we need to find the roots. Use the quadratic equation:

From here we split off into two roots, one where we add and one where we subtract:

and

Do both of these roots satisfy ? Yes.

Then we move on to our next question: Does the graph shift from negative to positive at either of these roots? Yes. When , the graph shifts from negative to positive.

Therefore, the local minimum is at .

To find the minimum we need to look at the first derivative.

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, .

Now we need to find the roots of the derivative. 

Does  cross the -axis? Yes, it crosses at .

Our next question is, "Does the graph change from negative to positive at that point?" Yes. This means that our parent function has shifted from decreasing to increasing at that point. 

Therefore, this will be our minimum.

A local minimum occurs when a graph "bottoms out" -- it has been decreasing, it slows down, stops, and then begins to increase. At that point when it switches from decreasing to increasing, our first derivative should move from negative to positive. Start by finding the first derivative and then see if that happens.

To find the first derivative of , we can use the power rule.

The power rule states that we multiply each variable by its current exponent and then lower the exponent of each variable by one.

Since , we're going to treat  as .

Anything times zero is zero, so our final term , regardless of the power of the exponent.

Simplify what we have.

Our first derivative, then, is .

Graph the equation . Does it move from negative to positive when ? Yes it does. Therefore, that zero point will be our minimum.

The zero occurs when . Therefore the minimum of our original graph is .

To find the maximum, we need to look at the first derivative. 

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat  as  since anything to the zero power is one.

Notice that  since anything times zero is zero.

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Because we want the MINIMUM, we want to see where the derivative changes from negative to positive.

Notice that  has a root when . In fact, it changes from negative to positive at that particular point. This is the local minimum in the interval .

Substitute  to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as , .)

, so

, which is therefore the correct answer choice.

You can substitute  to write this as:

Note that as , 

, since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

, which is the correct choice.