AP Calculus AB Exam Question:

Let f(x) be a function defined by the antiderivative:

f(x) = ∫[0 to x] e^(3t)/(1+e^(2t)) dt

a) Determine the equation of the tangent line to the curve y = f(x) at the point where x = 0.

b) Find the x-coordinate of the point on the curve y = f(x) where the tangent line is horizontal.

Solution:

a) To determine the equation of the tangent line at the point (0, f(0)), we need to find both the value of f(0) and the derivative of f(x) at x = 0.

First, we find the value of f(0) by substituting x = 0 into the given antiderivative:

f(0) = ∫[0 to 0] e^(3t)/(1+e^(2t)) dt = 0

Next, we find the derivative of f(x) using the Fundamental Theorem of Calculus:

f'(x) = d/dx ∫[0 to x] e^(3t)/(1+e^(2t)) dt

Since the upper limit is dependent on x, we apply the Leibniz Integral Rule to differentiate the integral:

f'(x) = e^(3x)/(1+e^(2x)) * d/dx(x) - e^(3(0))/(1+e^(2(0))) * d/dx(0) = e^(3x)/(1+e^(2x))

To find the derivative at x = 0, we substitute x = 0 into f'(x):

f'(0) = e^(3(0))/(1+e^(2(0))) = 1/(1+1) = 1/2

Therefore, the equation of the tangent line at x = 0 is y = f'(0)(x - 0) + f(0), which simplifies to:

y = 1/2x

b) To find the x-coordinate of the point where the tangent line is horizontal, we set the derivative equal to zero and solve for x:

f'(x) = 0

e^(3x)/(1+e^(2x)) = 0

Since the exponential function e^k is never equal to zero for any real number k, the denominator (1+e^(2x)) cannot be zero.

Therefore, there is no value of x where the tangent line is horizontal.

Hence, the x-coordinate of the point where the tangent line is horizontal does not exist.