Pre-Calculus - Week 1

Homework

Check each answer individually. Use the hint if you get stuck, then ask the tutor.

Problem 1 · easy

If f(x) = 5x - 2, find f(3).

Problem 2 · easy

If g(x) = x^2 - 4x, find g(-1).

Problem 3 · medium

If f(x) = 2x + 3, find f(a - 1) and simplify.

Problem 4 · easy

Find the domain of f(x) = 1/(x + 5). Write in interval notation.

Problem 5 · easy

Find the domain of g(x) = sqrt(3x - 9). Write in interval notation.

Problem 6 · medium

Find the domain of h(x) = sqrt(x + 4) / (x - 2). Write in interval notation.

Problem 7 · easy

Is y = x^2 a function? Justify using the vertical line test.

Problem 8 · easy

Describe the transformation: g(x) = (x - 4)^2 + 1 relative to f(x) = x^2.

Problem 9 · medium

Describe: g(x) = -sqrt(x) + 3 relative to f(x) = sqrt(x).

Problem 10 · medium

Determine if f(x) = x^4 - 2x^2 is even, odd, or neither.

Problem 11 · medium

Determine if f(x) = x^3 + x is even, odd, or neither.

Problem 12 · medium

Determine if f(x) = x^2 + x is even, odd, or neither.

Problem 13 · medium

Find the average rate of change of f(x) = x^2 on [2, 5].

Problem 14 · easy

Find the average rate of change of f(x) = 2x - 3 on [1, 4].

Problem 15 · medium

If f(x) = 3x and g(x) = x - 1, find f(g(x)).

Problem 16 · medium

If f(x) = x^2 and g(x) = 2x + 1, find g(f(x)).

Problem 17 · medium

Evaluate the piecewise function: f(x) = {x + 1 if x < 2; x^2 if x >= 2} at x = -3.

Problem 18 · easy

Evaluate the same piecewise function at x = 3.

Problem 19 · medium

Identify the toolkit function: f(x) = 1/x. What are its asymptotes?

Problem 20 · medium

Write the equation of the absolute value function shifted left 3 and down 5.

Problem 21 · medium

Write the equation of the parabola with vertex (2, -7) opening upward.

Problem 22 · medium

Find the range of f(x) = x^2 - 3.

Problem 23 · challenge

If f(x) = sqrt(x) and g(x) = x - 4, find the domain of f(g(x)).

Problem 24 · medium

Evaluate: if h(x) = 2x^2 - 3x + 1, find h(0) + h(1).

Problem 25 · challenge

Describe all transformations: g(x) = 2|x - 1| + 3 relative to f(x) = |x|.