Question

How many fixed points are there for the function f(x) = x^5?

Answer 1

Using the fixed point concept, it is found that f(x) has 3 real fixed points.

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The fixed points of a function f(x) are the values of x for which:

$f(x) = x$

In this question:

$f(x) = x^5$

Then

$x = x^5$

$x^5 - x = 0$

$x(x^4 - 1) = 0$

Applying subtraction of perfect squares, $x^4 - 1 = (x^2 - 1)(x^2 + 1)$

Again, $x^2 - 1 = (x - 1)(x + 1)$

Then

$x(x^4 - 1) = 0$

$x(x - 1)(x + 1)(x^2 + 1) = 0$

There are 3 real fixed points, $x = 0, x = 1$ and $x = -1$.

A similar problem is given at brainly.com/question/22560442