e^3 x =
3 x = ln
x = ln ≈ −.135
Remember to round all answers to three decimal places on the AP Exam.
- C First, let’s find the derivative: = 3x^2 + 24x + 15.
Next, set the derivative equal to zero and solve for x.
3 x^2 + 24x + 15 = 0
x^2 + 8x + 5 = 0
Using the quadratic formula (or a calculator), we get
x = ≈ −0.683, −7.317
Let’s use the second derivative test to determine which is the maximum. We take the second
derivative and then plug in the critical values that we found when we set the first derivative
equal to zero. If the sign of the second derivative at a critical value is positive, then the curve
has a local minimum there. If the sign of the second derivative is negative, then the curve has a
local maximum there.
The second derivative is = 6x + 24. The second derivative is negative at x = −7.317, so
the curve has a local maximum there.
- B The formula for the perimeter of a square is P = 4s, where s is the length of a side of the
square.
If we differentiate this with respect to t, we get = 4 . We plug in = 0.4 to get =
4(0.4) = 1.6.
The formula for the area of a square is A = s^2 . If we solve the perimeter equation for s in terms
of P and substitute it into the area equation, we get
