CLEP College Algebra Practice Test

$4x^{2}$ - 12xy + $9y^{2}$ is correct because the binomial theorem states that $(a - b)^{2}$ = $a^{2}$ - 2ab + $b^{2}$, so $(2x - 3y)^{2}$ = $(2x)^{2}$ - 2*(2x)*(3y) + $(3y)^{2}$ = $4x^{2}$ - 12xy + $9y^{2}$.

6 is correct because the constant term of $(x + 1/x)^{4}$ occurs when two terms contribute x and the other two contribute 1/x, so the constant term is [4!/(2!(4-2)!)] = 6.

$40x^{3}$ is correct because the binomial theorem states that the kth term of $(a + b)^{n}$ is [n!/(k!(n-k)!)]$a^{n-k}$$b^{k}$, and for the 3rd term of $(x + 2)^{5}$, k = 2, so the term is [5!/(2!(5-2)!)]$x^{5-2}$$2^{2}$ = $40x^{3}$.

x < -3 is correct because subtracting 2x from both sides of x > 3 + 2x gives -x > 3, and then multiplying by -1 and reversing the inequality gives x < -3, applying the rules of solving linear inequalities.

Inverse functions satisfy the condition that their composition equals the input, meaning f(g(x)) = x and g(f(x)) = x. Since f(g(x)) and g(f(x)) both simplify to x, f and g are indeed inverse functions.

(x+2)/3 is correct because to find the inverse of f(x) = 3x - 2, we swap x and y to get x = 3y - 2, then solve for y, which gives y = (x+2)/3.

x/2 is correct because to find the inverse of f(x) = 2x, we swap x and y to get x = 2y, then solve for y, which gives y = x/2.

The domain is (1, ∞) because the expression under the square root must be positive and the denominator cannot be zero, so x - 1 > 0, which means x > 1.

The range is (-∞, ∞) because the linear function can take on any real value as x varies.

The domain is [-3, ∞) because the expression under the square root must be non-negative, so x + 3 ≥ 0, which means x ≥ -3.