Consider the mapping f : R → R given by f(x) = 4x ⁄ (x² + 1). Sketch the graph of f. Find an interval A = [—k, k] on the x-axis such that (a) {f(x) |x ∈ A} = Im f; (b) g : A → Im f given by g(a) =f(a) for every a ∈ A is a bijection. Obtain a formula for g^-1 : Im f → A

Question

Consider the mapping [latex]f\colon\mathbb{R} o\mathbb{R}[/latex] given by

[latex]f(x)={\frac{4x}{(x^{2}+1)}}\ .[/latex]

Sketch the graph of f. Find an interval A = [—k, k] on the x-axis such that

(a) {f(x) |x ∈ A} = Im f;

(b) g : A → Im f given by g(a) =f(a) for every a ∈ A is a bijection.

Obtain a formula for [latex]g^{-1}[/latex] : Im f → A

Accepted Answer

Use calculus to determine the graph of f (Fig. S3.39). From this graph we see that Im f = [—2, 2] and that the required interval A is [—1, 1]. To find [latex]g^{-1}[/latex] we set [latex]y=4x/(x^{2}+1),[/latex] so that [latex]y x^{2}-4x+y=0,[/latex] and solve this quadratic

for x in terms of y. We get

[latex]x=\left\{\begin{array}{l}\frac{4 \pm \sqrt{ }\left(16-4 y^2 ight)}{2 y}=\frac{2 \pm \sqrt{ }\left(4-y^2 ight)}{y} ext { if } y eq 0 \0 \quad ext { if } y=0 .\end{array} ight.[/latex]

This suggests either that

[latex]g^{-1}(y)=\left\{\begin{array}{ccc}\frac{2+\sqrt{ }\left(4-y^2 ight)}{y} & ext { if } & y eq 0 \0 & ext { if } & y=0,\end{array} ight.[/latex]

or that

[latex]g^{-1}(y)=\left\{\begin{array}{ccc}\frac{2-\sqrt{ }\left(4-y^2 ight)}{y} & ext { if } & y eq 0 ; \0 & ext { if } & y=0 .\end{array} ight.[/latex]

The first possibility is excluded since, for [latex]y\in[-1,1],[/latex] we have

[latex]{\frac{2+{\sqrt{(4-y^{2})}}}{y}} ot\in[-2,2].[/latex]

Question 3.32: Consider the mapping f : R → R given by f(x) = 4x ⁄ (x² + 1......

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