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Another Question, But This One is About Proving a Function is One-to-One (PLEASE HELP)

AutGuy98

New member
Sep 11, 2019
20
Hey guys, got another question for you to look at and hopefully help me out on.

For each of the following functions, prove that f is one to one on E and find a formula for the inverse function f-1.

(a) f(x)=x2+3x-6 and E=[-3/2,infinity).

(b) f(x)=((x)/(x2+1)) and E=[-1,1].

Please help me here, mainly with proving that each of the two functions are one-to-one, since that's what I'm mainly having trouble with. I did manage to calculate the formulas for the inverse functions of both, but there are plus or minuses involved (since I used the Quadratic Formula to solve for them) and that is where I've stopped with that. However, I need to choose only one of the solutions (either positive or negative) for each function and I'm not sure how to do that. Could someone please help me show that each function is one-to-one and help me choose between a positive and negative formula for each inverse function? These are what I have for them:

(a) f-1(x)=-3/2+((sqrt(4x+33))/(2)) and f-1(x)=-3/2-((sqrt(4x+33))(2)).
(b) f-1(x)=((1)/(2x))+((sqrt(1-4x2))/(2x)) and f-1(x)=((1)/(2x))-((sqrt(1-4x2))/(2x))

Again, thank you guys for your help before, I really do appreciate all of it, and I hope you can read this over quickly enough to get back to me in time. Thank you in advance for doing so.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
644
For each of the following functions, prove that f is one to one on E and find a formula for the inverse function f-1.

(a) f(x)=x2+3x-6 and E=[-3/2,infinity).

(b) f(x)=((x)/(x2+1)) and E=[-1,1].
Let $a \text{ and } b$ be two elements in the domain of a function $f$. The function, $f$ , is one-one if $f(a)=f(b) \implies a = b$

For $f(x) = x^2+3x-6$ ... set $f(a)=f(b)$

$a^2 + 3a - 6 = b^2 + 3b - 6$

$a^2+3a = b^2+3b$

$a^2-b^2 = 3b-3a$

$(a-b)(a+b) = 3(b-a)$

$(a-b)(a+b) + 3(a-b) = 0$

$(a-b)[(a+b)+3] = 0 \implies a=b \text{ or } a+b=-3$

If $a=b$, we're done. If $a,b \in \left[-\dfrac{3}{2},\infty \right)$, then $a=b=-\dfrac{3}{2}$. If $a \ne b$, then $a + b \ne -3 \implies (a+g)+3 \ne 0$

Inverse, solve for $x$ ...

$y = x^2+3x-6$

$y+6 = x^2+3x+\dfrac{9}{4}$

$y+6 = \left(x+\dfrac{3}{2}\right)^2$

$\sqrt{y+6} = \left|x+\dfrac{3}{2}\right|$

case 1 ... $x+\dfrac{3}{2} \ge 0 \implies x+\dfrac{3}{2} = \sqrt{y+6} \implies x = \sqrt{y+6} - \dfrac{3}{2}$

note the domain of the inverse requires $y \ge -6 \implies x \ge -\dfrac{3}{2}$

case 2 ... $x+\dfrac{3}{2} < 0 \implies x+\dfrac{3}{2} = -\sqrt{y+6} \implies x = -\sqrt{x+6} - \dfrac{3}{2}$

note the domain of the inverse requires $y \ge -6 \implies x < -\dfrac{3}{2}$

note that case 1 follows the given domain restriction for $f \implies f^{-1}(x) = \sqrt{x+6} - \dfrac{3}{2}$


------------------------------------------------------------------------------------------------------------------

$f(x) = \dfrac{x}{x^2+1}$

$\dfrac{a}{a^2+1} = \dfrac{b}{b^2+1}$

$a(b^2+1) = b(a^2+1)$

$ab^2+a = a^2b+b$

$ab^2-a^2b = b-a$

$ab(b-a) = b-a$

$ab(b-a)-(b-a) =0$

$(b-a)(ab-1) = 0 \implies b=a \text{ or } ab=1$

If $b=a$, we're done. $ab=1 \implies a = b= \pm 1$ for any $a,b \in [-1,1]$

Inverse, solve for $x$ ...

$y=\dfrac{x}{x^2+1}$ note $y = f(0) = 0$

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

$yx^2 - x + y = 0$

$x = \dfrac{1 \pm \sqrt{1 - 4y^2}}{2y}$ for $y \in \left[-\dfrac{1}{2} , 0\right) \cup \left(0, \dfrac{1}{2}\right]$

note $f\left(\dfrac{1}{2}\right) = \dfrac{\frac{1}{2}}{\frac{1}{4}+1} = \dfrac{2}{5} \implies f^{-1}\left(\dfrac{2}{5}\right)=\dfrac{1}{2} \implies \dfrac{1}{2} = \dfrac{1 - \sqrt{1-\frac{16}{25}}}{\frac{4}{5}} $, and ...

$f\left(-\dfrac{1}{2}\right) = \dfrac{-\frac{1}{2}}{\frac{1}{4}+1} = -\dfrac{2}{5} \implies f^{-1}\left(-\dfrac{2}{5}\right)=-\dfrac{1}{2} \implies -\dfrac{1}{2} = \dfrac{1 - \sqrt{1-\frac{16}{25}}}{-\frac{4}{5}} $

therefore, $x$ as a function of $y$ is

$x = \left\{\begin{matrix}
0 & y=0\\
\dfrac{1-\sqrt{1-4y^2}}{2y} & y \in \left[-\dfrac{1}{2} , 0\right) \cup \left(0, \dfrac{1}{2}\right]
\end{matrix}\right.$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
For (a) the first thing I would do is "complete the square".
[tex]y= x^2+ 3x- 6= x^2+ 3x+ 9/4- 9/4- 24/4= (x+ 3/2)^2- 33/4[/tex].

The graph of that is a parabola, opening upward, with vertex at (-3/2, -33/4).
For x> -3/2 we have the single branch of the parabola so this function is "one to one".

Since [tex]y= (x+ 3/2)^2- 33/4[/tex], [tex](x+ 3/2)^2= (y+ 33/4)[/tex], [tex]x+ 3/2= \sqrt{y+ 33/4}[/tex] (since x> -3/2, x+ 3/2 is positive and we must take the positive root) so [tex]x= -3/2+ \sqrt{y+ 33/4}[/tex]. Switch "x" and "y" to get the inverse function.
 
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