Inverse Function

Aladdin

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Given the function f defined by f(x)=x1+x2\displaystyle f(x) = \frac{x}{\sqrt{1+x^2}}

1) Show that f admits an inverse over R .

2) Determine the Domain of the inverse .

3) Express the inverse function .

Work :

A function admits an inverse in it's domain if it is continuous , defined and strictly monotonic .

f(x) is always cont. def. in R

f(x) can be written (x)(1+x^2)^-1/2

f'(x) = (x)(-1/2)(1+x^2)(2x) + (1)(1+x^2)^-1/2

f'(x) = (x^2)(1+x^2) + 1/sqrt{1+x^2)

f'(x) is always positive on R . So it's stricly increasing .

So f(x) admits an invers .

2) Domain of invers = range f(x)
R= ]-inf , +inf[
f^-1 = f(R) = ]f(-inf) , f(+inf)[ = ] ? , ? [

I think there's a trick to be done here . I've took x^2 as a common factor but still ....

3)f^-1 = ?
y=x/sqr{1+x^2}

y^2=x^2/1+x^2

y^2*(1+x^2) = x^2

y^2 + y^2*x^2 = x^2

y^2 = x^2(-y^2+1)

x^2= y^2/(-y^2+1)

x = +- y(/sqr{1}{-y^2+1) ....

Thanks in advance .
 
The range of f will be (m, M), where m is the horizontal asymptote as x approaches -inf, M, similarly for x -> +inf. This works only in this case, and note they are open intervals as you showed the function is always monotonic.
 
daon said:
The range of f will be (m, M), where m is the horizontal asymptote as x approaches -inf, M, similarly for x -> +inf. This works only in this case, and note they are open intervals as you showed the function is always monotonic.

Thank you doan , but what's the value of m & M .
 
f(x) = y = x(1+x2)\displaystyle f(x) \ = \ y \ = \ \frac{x}{\sqrt(1+x^{2})}

Change: x = y(1+y2)\displaystyle Change: \ x \ = \ \frac{y}{\sqrt(1+y^{2})}

Ergo, x2 = y21+y2, 1x2 = 1+y2y2\displaystyle Ergo, \ x^{2} \ = \ \frac{y^{2}}{1+y^{2}}, \ \frac{1}{x^{2}} \ = \ \frac{1+y^{2}}{y^{2}}

1x2 = 1y2+1, 1y2 = 1x21\displaystyle \frac{1}{x^{2}} \ = \ \frac{1}{y^{2}}+1, \ \frac{1}{y^{2}} \ = \ \frac{1}{x^{2}}-1

1y2 = 1x2x2, y2 = x21x2\displaystyle \frac{1}{y^{2}} \ = \ \frac{1-x^{2}}{x^{2}}, \ y^{2} \ = \ \frac{x^{2}}{1-x^{2}}

y = f1(x) = x(1x2), 1 < x < 1\displaystyle y \ = \ f^{-1}(x) \ = \ \frac{x}{\sqrt(1-x^{2})}, \ -1 \ < \ x \ < \ 1

Check: f[f1(x)] = x = f1[f(x)], see graph below.\displaystyle Check: \ f[f^{-1}(x)] \ = \ x \ = \ f^{-1}[f(x)], \ see \ graph \ below.

[attachment=0:222731xo]abc.jpg[/attachment:222731xo]
 

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Thank you Gleen ...

Can you please , clarify number 2) for me .

Lim ( (x)/sqrt(1+x^2) as x tends to infinity ...
 
limxx(1+x2) = limxxx2 = limxxx\displaystyle \lim_{x\to\infty}\frac{x}{\sqrt(1+x^{2})} \ = \ \lim_{x\to\infty} \frac{x}{\sqrt x^{2}} \ = \ \lim_{x\to\infty}\frac{x}{|x|}

= limxxx = 1\displaystyle = \ \lim_{x\to\infty}\frac{x}{x} \ = \ 1

Is this what you wanted?\displaystyle Is \ this \ what \ you \ wanted?
 
BigGlenntheHeavy said:
limxx(1+x2) = limxxx2 = limxxx\displaystyle \lim_{x\to\infty}\frac{x}{\sqrt(1+x^{2})} \ = \ \lim_{x\to\infty} \frac{x}{\sqrt x^{2}} \ = \ \lim_{x\to\infty}\frac{x}{|x|}

= limxxx = 1\displaystyle = \ \lim_{x\to\infty}\frac{x}{x} \ = \ 1

Is this what you wanted?\displaystyle Is \ this \ what \ you \ wanted?

Yes ~ Thank you Gleen ...

Aha , So as x tends to -inf limit is -1

& as x tends to +inf limit is 1 .
 
Right, see below:\displaystyle Right, \ see \ below:

limxx(1+x2) = limxxx2\displaystyle \lim_{x\to-\infty}\frac{x}{\sqrt(1+x^{2})} \ = \ \lim_{x\to-\infty}\frac{x}{\sqrt x^{2}}

= limxxx = limxxx = 1\displaystyle = \ \lim_{x\to-\infty}\frac{x}{|x|} \ = \ \lim_{x\to-\infty}\frac{x}{-x} \ = \ -1

See graph below:\displaystyle See \ graph \ below:

[attachment=0:x8yqqpjc]def.jpg[/attachment:x8yqqpjc]
 

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