Final answer:
To determine the values of p and q for which the function f(x) = x³ + px + 2q is injective, we need to find when the function satisfies the condition of injectivity, which is that for any two distinct inputs x₁ and x₂, the outputs f(x₁) and f(x₂) are also distinct. The values of p and q that make the function injective are those that satisfy the equation x₁² + x₁*x₂ + x₂² + p = 0.
Step-by-step explanation:
To determine the values of p and q for which the function f(x) = x³ + px + 2q is injective, we need to find when the function satisfies the condition of injectivity, which is that for any two distinct inputs x₁ and x₂, the outputs f(x₁) and f(x₂) are also distinct.
This means that if f(x₁) = f(x₂), then x₁ = x₂. In other words, the function is injective if and only if it has no repeated outputs (no two different values of x map to the same value of f(x)).
In the case of the function f(x) = x³ + px + 2q, we can set the outputs equal to each other and solve for x to determine the values of p and q.
Let's set f(x₁) = f(x₂) and solve for x:
x₁³ + p*x₁ + 2q = x₂³ + p*x₂ + 2q
x₁³ - x₂³ + p*x₁ - p*x₂ = 0
(x₁ - x₂)(x₁² + x₁*x₂ + x₂²) + p(x₁ - x₂) = 0
(x₁ - x₂)(x₁² + x₁*x₂ + x₂² + p) = 0
Since x₁ and x₂ are distinct, the only way for the equation to be true is if the factor x₁² + x₁*x₂ + x₂² + p is equal to zero. This gives us the condition for the function to be injective:
x₁² + x₁*x₂ + x₂² + p = 0
Therefore, the values of p and q that make the function f(x) = x³ + px + 2q injective are those that satisfy the equation x₁² + x₁*x₂ + x₂² + p = 0.