Question

How many fixed points are there for the function f(x)=x^4

Answer 1

Answer:

What is this?

Step-by-step explanation:

Answer 2

The fixed points in the function f(x) = x^4 are:

x = 0, x = 1, x = -0.5 - 0.866i, and x = -0.5 +0.866i

The given function is:

$f(x) = x^4$

The fixed points of a function f(x) is determined by solving f(x) - x = 0

$x^4 - x = 0\\x(x^3-1) = 0\\x(x-1)(x^2+x+1) = 0$

Equate each of the terms to zero:

x  =  0

x  - 1  =  0

x  =  1

$x^2+x+1 = 0\\\\x = \frac{-b\pm \sqrt{b^2-4ac} }{2a} \\\\x = \frac{-1\pm \sqrt{1^2-4(1)(1)} }{2(1)} \\\\x = \frac{-1\pm \sqrt{-3} }{2}\\\\x= \frac{-1\pm i\sqrt{3} }{2}\\\\x = \frac{-1 \pm i\sqrt{3} }{2}\\\\x= -0.5\pm0.866i\\\\x_1 = -0.5-0.8661\\\\x_2 = -0.5+0.866i$

Therefore, the fixed points in the function f(x) = x^4 are:

x = 0, x = 1, x = -0.5 - 0.866i, and x = -0.5 +0.8661

Learn more here: https://brainly.com/question/14785053