AP Calculus AB Exam Question:

Let f(x) = (3x^2 + 4x) / (2x - 1).

a) Find f'(x) using the quotient rule. b) Use f'(x) to find the equation of the tangent line to the curve y = f(x) at the point where x = 2.

Answer:

a) To find f'(x) using the quotient rule, we need to differentiate the numerator and denominator separately and apply the quotient rule formula.

Let's begin by differentiating the numerator:

f'(x) = (d/dx)[3x^2 + 4x]

Using the power rule, we can find the derivative of each term separately:

= (d/dx)(3x^2) + (d/dx)(4x) = 6x + 4

Now, let's differentiate the denominator:

(d/dx)(2x - 1) = 2

Applying the quotient rule formula, we have:

f'(x) = (numerator's derivative * denominator - numerator * denominator's derivative) / (denominator)^2 = (6x + 4)(2x - 1) - (3x^2 + 4x)(2) / (2x - 1)^2 = (12x^2 - 2x + 8x - 4 - 6x^2 - 8x) / (2x - 1)^2 = (6x^2 - 2x - 4) / (2x - 1)^2

Therefore, f'(x) = (6x^2 - 2x - 4) / (2x - 1)^2.

b) To find the equation of the tangent line to the curve y = f(x) at the point where x = 2, we need to find the corresponding y-coordinate and slope at that point.

Using the equation y = f(x), substitute x = 2:

y = f(2) = (3(2)^2 + 4(2)) / (2(2) - 1) = (12 + 8) / (4 - 1) = 20 / 3

Thus, the point of tangency is (2, 20/3).

Now, substitute x = 2 into f'(x) to find the slope at x = 2:

f'(2) = (6(2)^2 - 2(2) - 4) / (2(2) - 1)^2 = (24 - 4 - 4) / (4 - 1)^2 = 16/9

Therefore, the slope of the tangent line is 16/9.

Using the point-slope form of a line, we can write the equation of the tangent line as:

y - (20/3) = (16/9)(x - 2)

Simplifying the equation, we get:

9y - 60 = 16x - 32

Hence, the equation of the tangent line to the curve y = f(x) at the point where x = 2 is 16x - 9y - 28 = 0.