For the function $f:\lbrack 1,\infty) \rightarrow \lbrack 1,\infty)$ defined by $f(x) = (x - 1)^{4}

Question

For the function $f:\lbrack 1,\infty) \rightarrow \lbrack 1,\infty)$ defined by $f(x) = (x - 1)^{4} + 1$, among the two statements :

(I) The set $S = \left\{ x \in \lbrack 1,\infty):f(x) = f^{1}(x) \right\}$ contains exactly two elements, and

(II) The set $S = \left\{ x \in \lbrack 1,\infty):f(x) = f^{1}(x + 1) \right\}$ is an empty set,

Accepted Answer

Sol. [Key Concept: For an increasing $f$, $f = f^{- 1}$ meets on $y = x$; solve $(x{- 1)}^{4} + 1 = x$ and test each statement.]$${f(x) = (x - 1)^{4} + 1 }{f^{'}(x) = 4(x - 1)^{3};f^{'}(x) \geq 0 }$$$	herefore f(x)$ is increasing$${	herefore(x - 1)^{4} + 1 = x }{(x - 1)\left\lbrack (x - 1)^{3} - 1 ightbrack = 0 }$$$x = 1,x = 2$ are two solutionsNow$${f^{- 1}(x) = (x - 1)^{1/4} + 1 }{f^{- 1}(x + 1) = x^{1/4} + 1 }{f^{- 1}(x + 1) = f(x) }{x^{1/4} + 1 = (x - 1)^{4} + 1 }{x = (x - 1)^{16} }$$Above equation has one solution using graph.

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