{Use of Tech} Fixed points of quadratics and quartics Let f(x) = ax(1 -x), where a is a real number and 0 ≤ a ≤ 1. Recall that the fixed point of a function is a value of x such that f(x) = x (Exercises 48–51).​​ 


d. Find the number and location of the fixed points of g for a = 2, 3, and 4 on the interval 0 ≤ x ≤ 1. 

Verified step by step guidance

1. Start by setting the function f(x) = ax(1 - x) equal to x to find the fixed points: ax(1 - x) = x.
2. Rearrange the equation to form a standard polynomial equation: ax(1 - x) - x = 0, which simplifies to ax - ax^2 - x = 0.
3. Combine like terms to get a quadratic equation in the form of -ax^2 + (a - 1)x = 0.
4. Factor the quadratic equation: x(-ax + (a - 1)) = 0, which gives you one fixed point at x = 0 and another at x = (a - 1)/a.
5. Evaluate the second fixed point for a = 2, 3, and 4 to determine if it lies within the interval 0 ≤ x ≤ 1.