Critical Points of Implicit Relations (College Board AP® Calculus AB)

Consider the function whose curve is given by the equation for .

(a) For and , find the coordinates of the point where the tangent to the curve is horizontal.

Answer:

We need to find where the derivative is equal to zero, as the slope of the tangent is zero at that point

Differentiate both sides of the equation with respect to

Use the product rule for the right hand side, and don't forget to use the chain rule when differentiating with respect to

Rearrange for

Set this equal to zero to find the critical point

We also need to be careful with the denominator, so that the derivative is not undefined (from dividing by zero)

and

Solve to find , and use the fact that we know , so will have the same solutions as

We were told in the question that

Find the value by substituting into

, so

Check this satisfies

The point where the tangent to the curve is horizontal is

(b) Determine whether the curve has a relative minimum, a relative maximum, or neither at the point found in part (a). Justify your answer.

Answer:

We need to check if the second derivative is positive or negative at

Write down the first derivative, and then differentiate both sides with respect to

Use the quotient rule, with and

Differentiating is relatively straightforward using the chain rule

Differentiating will require the product rule

Use different variables than and so you don't get confused, e.g. and

Apply the quotient rule,

This is quite a tricky expression, but thankfully we do not need to rearrange it, or even substitute in the expression

We know that the point is and we know that at this point,

So we can substitute these values of , , and into the expression for

Use the sign of the second derivative to classify the nature of the critical point

At , the first derivative is zero, and the second derivative is negative.

Therefore it is a relative maximum.