AP Calculus AB Exam Question:

Consider the function f(x) defined by the equation:

f(x) = (6x^2 - 4x + 2)^(1/3)

1. Find the derivative of f(x).

2. Use the derivative to determine the slope of the tangent line to the graph of f(x) at the point where x = 1.

Solution:

To find the derivative of f(x), we will use the power rule and chain rule.

1. Using the power rule, we differentiate each term inside the parentheses:

f'(x) = (1/3) * (6x^2 - 4x + 2)^(-2/3) * (d/dx)(6x^2 - 4x + 2)

Now, we compute the derivative of the inside function using the product rule:

d/dx(6x^2 - 4x + 2) = 12x - 4

Substituting this value back into our derivative expression:

f'(x) = (1/3) * (6x^2 - 4x + 2)^(-2/3) * (12x - 4)

Simplifying further:

f'(x) = (12x - 4) / (3(6x^2 - 4x + 2)^(2/3))

2. To find the slope of the tangent line to the graph of f(x) at the point where x = 1, we substitute x = 1 into the derivative we found:

f'(1) = (12(1) - 4) / (3(6(1)^2 - 4(1) + 2)^(2/3))

Simplifying the expression:

f'(1) = 8 / (3(6 - 4 + 2)^(2/3))

f'(1) = 8 / (3(4)^(2/3))

Using the property of exponents that (a^b)^c = a^(b*c):

f'(1) = 8 / (3 * 4^(2/3))

Cube root of 4 (4^(1/3)) is equal to 2, so:

f'(1) = 8 / (3 * 2^2)

f'(1) = 8 / (3 * 4)

Therefore, the slope of the tangent line to the graph of f(x) at the point where x = 1 is 2/3.

Final Answer:

1. f'(x) = (12x - 4) / (3(6x^2 - 4x + 2)^(2/3))

2. The slope of the tangent line to the graph of f(x) at the point where x = 1 is 2/3.