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:
[tex]f(x) = x[/tex]
In this question:
[tex]f(x) = x^5[/tex]
Then
[tex]x = x^5[/tex]
[tex]x^5 - x = 0[/tex]
[tex]x(x^4 - 1) = 0[/tex]
Applying subtraction of perfect squares, [tex]x^4 - 1 = (x^2 - 1)(x^2 + 1)[/tex]
Again, [tex]x^2 - 1 = (x - 1)(x + 1)[/tex]
Then
[tex]x(x^4 - 1) = 0[/tex]
[tex]x(x - 1)(x + 1)(x^2 + 1) = 0[/tex]
There are 3 real fixed points, [tex]x = 0, x = 1[/tex] and [tex]x = -1[/tex].
A similar problem is given at https://brainly.com/question/22560442
