f(5) = 2(5) + 3 = 10 + 3 = 13.

f(2) = 2(2^2) - 3(2) + 1 = 8 - 6 + 1 = 3.

f(3) = 2^3 = 8.

f(-2) = 3(-2) - 4 = -6 - 4 = -10.

f(-2) = 3(-2) - 4 = -6 - 4 = -10.

To find the inverse, set y = 3x - 4, solve for x: x = (y + 4)/3, thus f^(-1)(x) = (x + 4)/3.

First, find f(1) = 3(1) - 5 = -2. Then, f(-2) = 3(-2) - 5 = -6 - 5 = -11.

f(π/2) = sin(π/2) = 1.

The vertex form is f(x) = (x + 1)^2, so the vertex is (-1, 0).

To find x-intercepts, set f(x) = 0: x^2 - 4 = 0, which gives x = ±2.

f(2) = 2^2 - 4*2 + 3 = 4 - 8 + 3 = -1.

(f ∘ g)(2) = f(g(2)) = f(2 + 1) = f(3) = 3^2 = 9.

f(-3) = (-3)^2 = 9.

f(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.

f(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.

f(2) = |2 - 2| = |0| = 0.

f(-3) = |-3| = 3.

g(-1) = 3(-1) + 2 = -3 + 2 = -1.

h'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so x = 1 and x = -1 are critical points.

h(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.

h(1) = 1^3 - 3*1 = 1 - 3 = -2.

The function has a minimum at x = 0, hence it is neither always increasing nor decreasing.

f(x) = (x - 2)^2 is the completed square form.

The function |x - 3| is continuous everywhere, including at x = 3.

The absolute value function has a minimum value of 0, hence the range is [0, ∞).

f(g(x)) = f(2x) = 2x + 1.

The function is undefined at x = 2, so the domain is all real numbers except 2.

The function is undefined at x = 3, so the domain is x ≠ 3.

The expression under the square root must be non-negative, so x - 1 >= 0, hence x >= 1.

To find the inverse, set y = 2x + 3, solve for x: x = (y - 3)/2.