Question

Please choose correct answer and explain solution quickly. This is

for numerical analysis
For what values of the positive constant " a " does the following function have a unique fixed on the interval [-2,2] ? f(x)=\frac{x^{2}-4}{a} a=2 a ≥ 3 a ≥ 4

Answer 1

The main answer to the question is "a ≥ 4."

To explain the solution, we need to find the values of "a" for which the function f(x) = (x^2 - 4)/a has a unique fixed point on the interval [-2,2].

To find the fixed points, we set f(x) equal to x:
(x^2 - 4)/a = x

To solve this equation, we multiply both sides by "a" to eliminate the denominator:
x^2 - 4 = ax

Rearranging the equation, we get:
x^2 - ax - 4 = 0

For the function to have a unique fixed point, the quadratic equation should have exactly one real root. This occurs when the discriminant (b^2 - 4ac) is equal to zero.

In our case, a = -a, b = -a, and c = -4. Plugging these values into the discriminant formula:
(-a)^2 - 4(1)(-4) = 0

Simplifying the equation:
a^2 + 16 = 0

Taking the square root of both sides:
a = ± √(-16)

Since "a" is a positive constant, there are no real solutions for "a" in this case.

Therefore, we can conclude that the correct answer is a ≥ 4.