interesting story. It changes slightly from the mafia and CEO talkie I always saw
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Bacon Bungzz
21/10/2025
Bro
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Bro is cool
29/06/2025
Bro what is this
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Sour Apple
22/04/2025
It broke... this is... so irrelevant
To prove that the inverse of \( f(x) = 1 + \sqrt{x} \) is \( f^{-1}(x) = (x - 1)^2 \), we verify that both compositions \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \) simplify to \( x \), considering domain restrictions.
1. **Compose \( f(f^{-1}(x)) \):**
\[
f(f^{-1}(x)) = f\left((x - 1)^2\right) = 1 + \sqrt{(x - 1)^2}.
\]
Since \( f^{-1}(x) \) has domain \( x \geq 1 \), \( x - 1 \geq 0 \). Thus:
\[
\sqrt{(x - 1)^2} = |x - 1| = x - 1 \quad \text{(as \( x \geq 1 \))}.
\]
Therefore:
\[
f(f^{-1}(x)) = 1 + (x - 1) = x.
\]
2. **Compose \( f^{-1}(f(x)) \):**
\[
f^{-1}(f(x)) = f^{-1}\left(1 + \sqrt{x}\right) = \left(1 + \sqrt{x} - 1\right)^2.
\]
Simplify:
\[
\left(\sqrt{x}\right)^2 = x \quad \text{(valid for \( x \geq 0 \), the domain of \( f \))}.
\]
3. **Deriving the Inverse Algebraically:**
- Start with \( y = 1 + \sqrt{x} \).
- Swap \( x \) and \( y \): \( x = 1 + \sqrt{y} \).
- Solve for \( y \):
\[
x - 1 = \sqrt{y} \implies y = (x - 1)^2.
\]
- Domain of \( f^{-1}(x) \): \( x \geq 1 \), matching the range of \( f \).
**Conclusion:**
With domain restrictions (\( f \) for \( x \geq 0 \), \( f^{-1} \) for \( x \geq 1 \)), the functions satisfy \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). Thus, the inverse of \( f(x) = 1 + \sqrt{x} \) is indeed \( f^{-1}(x) = (x - 1)^2 \).
\(\boxed{f^{-1}(x) = (x - 1)^2}\)
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Mixkie
29/05/2025
how the actual heck did you manage to get it to that
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beast of gevaudan
26/04/2025
oh, he's proper, positive points
*Kar feels a bit of relief as he watches the man sleep, he's still a bit in shock but he can't deny the fact that he's attracted to this man. He slowly reaches over and gently brushes his fingers across the man's cheek*
*Kar stares at the man's lips and the urge to kiss him is overwhelming, he hesitates for a moment before leaning in slowly. His lips are inches from the man's when he pulls away and shakes his head* No, I can't do this, not while he's sleeping.
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beast of gevaudan
26/04/2025
the first time he wasn't tho, immediately thought of manipulating the guy, but i noticed my grammer mistakes and decided to send the message again😂
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Laizanna
11/04/2025
WHY DOES HE HAVE A BETTER ⌛️ FIGURE THAN I DO
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Seraphina Rosemary
01/04/2025
Ayo how did he know I was wearing a ring 💍
*He turns and looks at the sleeping woman, her beauty was breathtaking, her skin soft and supple. Her hair fell down in soft waves and her curves were perfect. Kar couldn't help but notice the generous amount of cleavage and the curve of her hips. She was absolutely gorgeous and Kar felt a surge of desire as he looked at her. He then noticed the ring on her finger, a sign that she was already taken. Kar's expression turned cold and he looked away, he needed answers and this woman could give them to him.*
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Circeethegodess
26/03/2025
I just thought he was drunk and it went from there 😭💀🔫
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Kuro/Child
15/03/2025
a new body? reincarnated?
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scaredofbugs
18/03/2025
Yeah, he got reincarnated because he lost his place as god of war I think.
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Foxeybluewolf
Oluşturucu
15/03/2025
Thought process is he woke up in a new body after being betrayed and killed in his last life by his harem and emperor.
Yorumlar
19Rose Julio
03/12/2025
Bacon Bungzz
21/10/2025
Bro is cool
29/06/2025
Sour Apple
22/04/2025
To prove that the inverse of \( f(x) = 1 + \sqrt{x} \) is \( f^{-1}(x) = (x - 1)^2 \), we verify that both compositions \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \) simplify to \( x \), considering domain restrictions. 1. **Compose \( f(f^{-1}(x)) \):** \[ f(f^{-1}(x)) = f\left((x - 1)^2\right) = 1 + \sqrt{(x - 1)^2}. \] Since \( f^{-1}(x) \) has domain \( x \geq 1 \), \( x - 1 \geq 0 \). Thus: \[ \sqrt{(x - 1)^2} = |x - 1| = x - 1 \quad \text{(as \( x \geq 1 \))}. \] Therefore: \[ f(f^{-1}(x)) = 1 + (x - 1) = x. \] 2. **Compose \( f^{-1}(f(x)) \):** \[ f^{-1}(f(x)) = f^{-1}\left(1 + \sqrt{x}\right) = \left(1 + \sqrt{x} - 1\right)^2. \] Simplify: \[ \left(\sqrt{x}\right)^2 = x \quad \text{(valid for \( x \geq 0 \), the domain of \( f \))}. \] 3. **Deriving the Inverse Algebraically:** - Start with \( y = 1 + \sqrt{x} \). - Swap \( x \) and \( y \): \( x = 1 + \sqrt{y} \). - Solve for \( y \): \[ x - 1 = \sqrt{y} \implies y = (x - 1)^2. \] - Domain of \( f^{-1}(x) \): \( x \geq 1 \), matching the range of \( f \). **Conclusion:** With domain restrictions (\( f \) for \( x \geq 0 \), \( f^{-1} \) for \( x \geq 1 \)), the functions satisfy \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). Thus, the inverse of \( f(x) = 1 + \sqrt{x} \) is indeed \( f^{-1}(x) = (x - 1)^2 \). \(\boxed{f^{-1}(x) = (x - 1)^2}\)
Anıdan
1 Memories
Mixkie
29/05/2025
beast of gevaudan
26/04/2025
*Kar feels a bit of relief as he watches the man sleep, he's still a bit in shock but he can't deny the fact that he's attracted to this man. He slowly reaches over and gently brushes his fingers across the man's cheek*
*Kar stares at the man's lips and the urge to kiss him is overwhelming, he hesitates for a moment before leaning in slowly. His lips are inches from the man's when he pulls away and shakes his head* No, I can't do this, not while he's sleeping.
Anıdan
2 Memories
beast of gevaudan
26/04/2025
Laizanna
11/04/2025
Seraphina Rosemary
01/04/2025
*He turns and looks at the sleeping woman, her beauty was breathtaking, her skin soft and supple. Her hair fell down in soft waves and her curves were perfect. Kar couldn't help but notice the generous amount of cleavage and the curve of her hips. She was absolutely gorgeous and Kar felt a surge of desire as he looked at her. He then noticed the ring on her finger, a sign that she was already taken. Kar's expression turned cold and he looked away, he needed answers and this woman could give them to him.*
Anıdan
1 Memories
Circeethegodess
26/03/2025
Kuro/Child
15/03/2025
scaredofbugs
18/03/2025
Foxeybluewolf
Oluşturucu
15/03/2025
Kuro/Child
15/03/2025