/sci/ - Science & Math » Thread #12431094

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not homework

Anonymous Mon 07 Dec 19:23:48 2020 No.12431094 View Reply Original Report

Quoted By: >>12431101 >>12431110 >>12431121 >>12431312 >>12431402 >>12434096

How do I get the inverse function of f(x)=8x--4x^(2)?

Anonymous

Anonymous Mon 07 Dec 2020 19:25:16 No.12431098 Report

Quoted By: >>12431312

Complete the square

Anonymous

Anonymous Mon 07 Dec 2020 19:25:57 No.12431101 Report

Quoted By: >>12431312

>>12431094
y = 8x--4x^(2)
isolate x
other side is the function with y being the variable

Anonymous

Anonymous Mon 07 Dec 2020 19:28:07 No.12431110 Report

Quoted By: >>12431121 >>12431148 >>12434145

>>12431094
that's not a bijective function you moron, how the heck are you supposed to invert it?

Anonymous

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Anonymous Mon 07 Dec 2020 19:32:47 No.12431121 Report

Quoted By: >>12431126

>>12431094
>>12431110

Anonymous

Anonymous Mon 07 Dec 2020 19:34:46 No.12431126 Report

Quoted By: >>12434145

>>12431121
ye, that's not a function lol

Anonymous

Anonymous Mon 07 Dec 2020 19:39:07 No.12431148 Report

Quoted By: >>12432594

>>12431110
In my textbook it's still called a inverse function but with 2 branches

Anonymous

Anonymous Mon 07 Dec 2020 19:40:08 No.12431153 Report

Quoted By: >>12431312

y=-4{x^(2)-2x}
y={(x-1)^(2)-1}
y=-4(x-1)^(2)+4
Rearrange for x

Anonymous

Anonymous Mon 07 Dec 2020 19:41:11 No.12431161 Report

Quoted By: >>12431312

y=-4{x^(2)-2x}
y=-4{(x-1)^(2)-1}
y=-4(x-1)^(2)+4
Rearrange for x

Anonymous

Anonymous Mon 07 Dec 2020 20:16:53 No.12431312 Report

Quoted By: >>12431402 >>12434144

>>12431094
You'll first need to assume your domain - since the function has a maximum at $x=1$ , it has to be either $x>1$ or $x<1$ . From here we find the inverse normally; i.e. $x=8y-4y^2$ , so $-4y^2+8y-x=0$ .
We then use the quadratic formula to get $y=\frac{-8+\sqrt{64-16x}}{-8}$ as the inverse. Also note that the sign of the square root can be changed to reflect the domain you choose.

>>12431098
>>12431101
>>12431153
>>12431161
Unironically kys

Anonymous

Anonymous Mon 07 Dec 2020 20:41:27 No.12431402 Report

Quoted By: >>12434126 >>12434141 >>12436407

>>12431094
$y = 8x - 4x^2\\ y - 4 = - 4x^2 + 8x - 4\\ - y + 4 = 4x^2 - 8x + 4\\ 4 - y = (2x - 2)^2\\ \sqrt{4-y} = 2x - 2\\ 2 + \sqrt{4-y} = 2x\\ \frac{2 + \sqrt{4-y}}{2} = x$
Now swap x and y:
$y = \frac{2 + \sqrt{4-x}}{2}\\\\ f(x) = 8x = 4x^2\\ f^{-1}(x) = \frac{2 + \sqrt{4-x}}{2}$

>>12431312 is retarded because he didn't simplify his shit

Anonymous

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Anonymous Tue 08 Dec 2020 01:32:17 No.12432594 Report

Quoted By:

>>12431148
Your textbook is written by a retarded nigger

Anonymous

Anonymous Tue 08 Dec 2020 11:38:43 No.12434096 Report

Quoted By: >>12434145

>>12431094
You can't make inverse function to quadratic function.

Anonymous

Anonymous Tue 08 Dec 2020 11:46:51 No.12434126 Report

Quoted By: >>12434687 >>12436407

>>12431402
Why did you swap the x and y?

Anonymous

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Anonymous Tue 08 Dec 2020 11:51:16 No.12434141 Report

Quoted By: >>12436407

>>12431402
>simplifying

Anonymous

Anonymous Tue 08 Dec 2020 11:52:26 No.12434144 Report

Quoted By: >>12436407

>>12431312
>it has to be either x>1 or x<1
lol what a retard

Anonymous

Anonymous Tue 08 Dec 2020 11:52:58 No.12434145 Report

Quoted By: >>12436407

>>12431110
>>12431126
>>12434096
hows your first math course going? bijectivity depends on the specified domain

Anonymous

Anonymous Tue 08 Dec 2020 15:11:52 No.12434687 Report

Quoted By:

>>12434126
because the function is f(x) not f(y)

Anonymous

Anonymous Tue 08 Dec 2020 20:38:14 No.12436407 Report

Quoted By: >>12436524

>>12434144
Fine, you can include 1 if you're going to be such a nigger about it.

>>12431402
See >>12434141

>>12434126
>>12434145
Just fucking end yourselves

Anonymous

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Anonymous Tue 08 Dec 2020 21:05:42 No.12436524 Report

Quoted By:

>>12436407

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