2 x − 2x − 2y +6y = 0
− 2x + 6y = 2y − 2x
Factor out the , and then isolate it.
(6y − 2x) = 2y − 2x
- D For an absolute maximum, find the first derivative and set it equal to zero to determine the
critical points: = 20x^4 − 20x = 0. The critical points are at x = 0 and x = 1. Find the second
derivative to determine which of these points are maxima and which are minima. Maxima are
located where the second derivative is negative; minima where the second derivative is
positive: =80x^3 − 20. From here, x = 0 corresponds with maximum points. Finally, plug
in this value into the original equation to determine which has a higher y-value. At x = 0, y =
−8. However, you must remember when determining absolute maxima and minima on a closed
interval to always check the endpoints. So, plug in x = −2 and x = 2 into the original equation
and determine if they have y-values that are more positive than at x = 0. At x = −2, y = −176,
and at x = 2, y = 80. Therefore, x = 2 is the absolute maximum.
- A Using implicit differentiation, find the first derivative of the equation and solve for : =
. Next, determine the second derivative, but don’t simplify the equation: =
. Finally, evaluate the first and second derivative at (−1,
−2). The first derivative at (−1, −2) is 2. The second derivative at (−1, −2) is .
