Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$.

Question

Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$.Find $N.$

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$\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$.Its graph look like the graph of $\frac{1}{x^2+1}$ but peak at $(1,1)$ buti do not know how does the graph of $f(f(x))$ behave and hence i am not able to find the number of points of intersection of $f(f(x))$ and $\frac{1}{x^2-2x+2}$.

Answer 1

Let $g(x)=f(f(x))$, then $g'(x)=f'(x)f'(f(x))\geq0$ since $f(x)$ is monotonic function.

Note that $x=1$ is the solution of $g(x)=f(f(x))=\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$

Now, since $\frac{1}{(x-1)^2+1}$ is strictly decresing function on interval $[1,\infty)$, and $g(x)$ is nondecreasing on that interval, the solution $x=1$ is unique. Hence, $N=1$.

Answer 2

Hint: If $f(x)$ is monotonic, then $f(f(x))$ is (possibly non-strictly) increasing, so…

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Full solution: Clearly if $f$ is increasing, so is $f\circ f$. Furthermore, if $f$ is decreasing, then $x \leq y$ implies $f(x) \geq f(y)$, which in turn implies $f(f(x)) \leq f(f(y))$. Either way, $f \circ f$ is increasing.

Since $f$ is $1$ at $x=1$, it follows that $f \circ f$ must have minimum $1$ at $x=1$. But it is easy to see that $(x^2-2x+2)^{-1}$ has maximum $1$ at $x=1$, thus $N=1$.

In particular, note that you don't need the hypothesis that $f$ is differentiable.