Problem 398 (difficulty: 2/10)

Are the following functions injective on \(\displaystyle [-1,1]\)?

a) \(\displaystyle \displaystyle f(x)=\frac{x}{x^2+1}\),    b) \(\displaystyle \displaystyle g(x)=\frac{x^{2}}{x^2+1}\).

Solution:

a) Let \(\displaystyle x\not=y\) and suppose that \(\displaystyle f(x)=f(y)\), i.e.,

\(\displaystyle \frac{x}{x^2+1}=\frac{y}{y^2+1}\ \Longrightarrow \ x(y^2+1)=y(x^2+1)\ \Longrightarrow \ x-y=(x-y)xy \ \Longrightarrow \ 1=xy,\)

since \(\displaystyle x-y\neq0\). On the other hand \(\displaystyle |x|,|y|\leq 1\), which can be satisfied only for \(\displaystyle x=y=\pm1\) but equality was not allowed. Therefore \(\displaystyle f(x)\) is injective on \(\displaystyle [-1,1]\).

b) \(\displaystyle g(1)=g(-1)\), therefore \(\displaystyle g(x)\) is not injective on \(\displaystyle [-1,1]\).

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