Function

[Aladdin]

Consider the function \(\displaystyle f(x) = x + \sqrt{x^2+1}\) , x belongs to R .

1) Show that f is strictly increasing over R .
2) Show that the two straight lines of equations y=0 and y=2x are asymptotes to (C).
3) Trace (C) .

4)a) Show that f admits an inverse function .
b) Trace the curve of the inverse on same system.
c) Find the equation of the inverse.

5) (D) is the domain limited by (C) , the axix x'x and the two straight lines of equations x=0 and x=1 .
Calculate the volume obtained by rotating (D) about x'x.

My work :

1) The derivative is \(\displaystyle f'(x) = 1 + \frac{x}{\sqrt{x^2+1}}\)
How can I prove it's +ve .
2) Lim[f(x)-0] as x tends to infinity and see if it equals to zero ~~~ Same as Lim[f(x)-2x] as x tends to infinity .

4) f(x) is continuous and defined for all x belongs to R .
f(x) is strictly increasing , monotomic --> f(x) admits an inverse .
b) Symmetry with respect to y=x.
c) \(\displaystyle x-y=\sqrt{x^2+1}\) Square bothe sides ... .. .

5) MmM . . .

Thanks in advance ,
Check my answers , please .

 

[soroban]

Hello, Aladdin!

Consider the function: .\(\displaystyle f(x) = x + \sqrt{x^2+1},\;\;x \in R.\)

1) Show that \(\displaystyle f\) is strictly increasing over \(\displaystyle R .\)


My work:

\(\displaystyle \text{1) The derivative is: }\: f'(x) \:=\:1 + \frac{x}{\sqrt{x^2+1}}\)
. . . How can I prove it's positive?

. . We start with: . . \(\displaystyle 0 \;<\;1\)

. . . . . . Add \(\displaystyle x^2\!:\quad x^2 \;<\;x^2+1\)

Take square root: . \(\displaystyle |x| \;<\;\sqrt{x^2+1}\)

. . . \(\displaystyle \text{Then: }\;\;\frac{|x|}{\sqrt{x^2+1}} \;<\,1\)


\(\displaystyle \text{This means: }\;-1 \;<\;\frac{x}{\sqrt{x^2+1}} \;<\;1\)

. . . \(\displaystyle \text{Add 1: }\;\;0 \;<\;1 + \frac{x}{\sqrt{x^2+1}} \;<\;2\)


\(\displaystyle \text{Therefore: }\;0 \:<\:f'(x)\:<\:2 \quad\hdots\; f\text{ is strictly increasing.}\)

 

[Aladdin]

Thank you , Soroban .

Nice trick . . .

 

[BigGlenntheHeavy]

\(\displaystyle 4c) \ f(x) \ = \ x+\sqrt{x^{2}+1}, \ f^{-1}(x) \ = \ \frac{x^{2}-1}{2x} \ | \ x \ > \ 0\)

\(\displaystyle See \ graph.\)

[attachment=0:39a2e7v0]SCREEN01.JPG[/attachment:39a2e7v0]

 

[Aladdin]

Thank you , Glenn ~

What about number 5 . . .

 

[BigGlenntheHeavy]

\(\displaystyle f'(x) \ = \ 1+\frac{x}{\sqrt{x^{2}+1}}, \ How \ can \ I \ prove \ it \ is \ positive?\)

\(\displaystyle There \ are \ only \ three \ possibilities \ for \ x, \ either \ x \ = \ 0. \ x \ > \ 0, \ or \ x \ < \ 0.\)

\(\displaystyle Hence, \ f'(x) \ > \ 0 \ ,positive \ by \ inspection, \ do \ you \ see \ why?\)

\(\displaystyle The \ only \ sticking \ point \ is \ when \ x \ < \ 0, \ but \ then \ \frac{x}{\sqrt{x^{2}+1}} \ range \ is \ (-1,1)\)

\(\displaystyle \ and \ 1 \ + \ a \ number \ in \ that \ range \ is \ always \ positive.\)

 

[Aladdin]

\(\displaystyle f'(x) \ = \ 1+\frac{x}{\sqrt{x^{2}+1}}, \ How \ can \ I \ prove \ it \ is \ positive?\)

\(\displaystyle There \ are \ only \ three \ possibilities \ for \ x, \ either \ x \ = \ 0. \ x \ > \ 0, \ or \ x \ < \ 0.\)

\(\displaystyle Hence, \ f'(x) \ > \ 0 \ ,positive \ by \ inspection, \ do \ you \ see \ why?\)

\(\displaystyle The \ only \ sticking \ point \ is \ when \ x \ < \ 0, \ but \ then \ \frac{x}{\sqrt{x^{2}+1}} \ is \ always \ less \ than \ 1\)

\(\displaystyle \ and \ 1 \ + \ a \ number \ less \ than \ 1 \ is \ always \ positive.\)

Yes , I can see why . . .

I'm going to use Soroban's justification , Thanks .

Glenn,What about the asymptotes . . .

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ x+\sqrt{x^{2}+1}\)

\(\displaystyle Now, \ as \ x \ increases \ without \ bounds, \ the \ 1 \ under \ the \ radical \ sign \ pails \ to \ insignficance, \ ergo\)

\(\displaystyle \ f(x) \ (oblique \ aysmptote) \ = \ x \ + \ \sqrt x^{2} \ = \ 2x\)

\(\displaystyle Same \ as \ x \ decreases \ without \ bounds, \ we \ get \ f(x) \ (horizontal \ aysmptote) \ = \ 0, \ -x \ + \ \sqrt(-x)^{2} \ =0.\)

\(\displaystyle See \ graph.\)

[attachment=0:1ekm3818]def.jpg[/attachment:1ekm3818]

 

[Aladdin]

I'm not getting it Glenn, Tomorrow I have a 3 hour math test & I think the teacher will put this question .

Why don't we use the limits . . .

 

[BigGlenntheHeavy]

\(\displaystyle \lim_{x\to\infty}f(x) \ = \ \infty, \ how \ can \ we \ use \ an \ aysmptote \ on \ infinity? \ We \ have \ to \ resort \ to \ plan \ B.\)

\(\displaystyle \lim_{x\to-\infty}f(x) \ = \ 0, \ here \ we \ can \ use \ the \ horizontal \ aysmptote.\)

 

[Aladdin]

\(\displaystyle \lim_{x\to\infty}f(x) \ = \ \infty, \ how \ can \ we \ use \ an \ aysmptote \ on \ infinity? \ We \ have \ to \ resort \ to \ plan \ B.\)

\(\displaystyle \lim_{x\to-\infty}f(x) \ = \ 0, \ here \ we \ can \ use \ the \ horizontal \ aysmptote.\)

\(\displaystyle \lim_{x\to\infty}f(x) \ - 2x \ = 0 \ that's \ how\ we \ prove\ that\ a\ line\ is \an\ oblique \ asymptote .\ . \. \ Right\)

 

[BigGlenntheHeavy]

\(\displaystyle Aladdin, \ let \ f(x) \ = \ x \ +\frac{1}{x}\)

\(\displaystyle Then, \ \lim_{x\to\infty}f(x) \ = \ \infty, \ and \ \lim_{x\to-\infty}f(x) \ = \ -\infty, \ right?\)

\(\displaystyle Now, \ does \ f(x) \ have \ a \ aysmptote? \ Yes, \ f(x) \ = \ x, \ an \ oblique \ aysmptote\)

\(\displaystyle See \ graph.\)

[attachment=0:1oi2lsxi]ghi.jpg[/attachment:1oi2lsxi]

 

[Aladdin]

For sure Glenn . . .

But what I'm linking for is : If the question is , Prove that y=x is an Oblique asymptote .

What you do is Find the limit of f(x)-x and see if it equals to zero . . .

Anyway , thanks again ~