Inverse Function

[dssHOPE]

I'm stumped, and it's probably something silly I'm missing. Can anyone tell me how to solve this?

 

[nycmathguy]

I'm stumped, and it's probably something silly I'm missing. Can anyone tell me how to solve this?
View attachment 17

f(x) = (x + 1)/(6x + 7)

Let y = f(x)

y = (x + 1)/(6x + 7)

x = (y+ 1)/(6y + 7)

Solve for y.

x(6y + 7) = y + 1

6xy + 7x = y + 1

6xy - y = -7x + 1

Factor out y.

y(6x - 1) = -7x+ 1

y = (-7x + 1)/(6x - 1)

f^(-1) x = (-7x + 1)/(6x - 1)

To find domain, set denominator to 0 and solve for x.
To find range, graph f^(-1) x using Desmos.

 

[MathLover1]

perfect!

domain: {x element R : x ≠1/6}
range:
{y element R : y≠-7/6}

 

[nycmathguy]

perfect!

domain: {x element R : x ≠1/6}
range:
{y element R : y≠-7/6}

Have you noticed that there are not too many participants here? Most of the threads date back 2 to 3 years and longer.

 

[MathLover1]

Have you noticed that there are not too many participants here? Most of the threads date back 2 to 3 years and longer.

yes, and there are not too many post either

 

[nycmathguy]

yes, and there are not too many post either

What do you really think? I can find another group for us. I want more participants. Understand? I love your dedication to mathematics, particularly my threads but in all honesty, it's not right, not fair for you to be the only reply, so to speak. In fact, there are different ways to solve math problems. Your method might be better than someone else's and vice-versa. I don't anyone is running thos site. I contacted administration yesterday. No reply thus far. This site is like a ghost town.

 

[Ian]

What do you really think? I can find another group for us. I want more participants. Understand? I love your dedication to mathematics, particularly my threads but in all honesty, it's not right, not fair for you to be the only reply, so to speak. In fact, there are different ways to solve math problems. Your method might be better than someone else's and vice-versa. I don't anyone is running thos site. I contacted administration yesterday. No reply thus far. This site is like a ghost town.

Sorry for the slow reply, that e-mail ended up in my spam folder for some reason - so I've only seen it when I saw your post.

Let's see if we can get this place active again, it has started to busy up with all of these recent posts .

 

[nycmathguy]

Sorry for the slow reply, that e-mail ended up in my spam folder for some reason - so I've only seen it when I saw your post.

Let's see if we can get this place active again, it has started to busy up with all of these recent posts .

My friend and I are the reason for the many threads. Where are the rest of the tutors? We need more people here. It's a good site. I joined to get help with my self-study of precalculus. I can answer math questions for grades 4 to 10.

 

[MathLover1]

My friend and I are the reason for the many threads. Where are the rest of the tutors? We need more people here. It's a good site. I joined to get help with my self-study of precalculus. I can answer math questions for grades 4 to 10.

It might be good idea for administrator to give opportunity to people to register as tutors, just like algebra.com did. If there are registered tutors here, students might come to seek help.

 

[MathLover1]

I'm stumped, and it's probably something silly I'm missing. Can anyone tell me how to solve this?
View attachment 17

given: f (x)=(x+1)/(6x+7)

find f'(x)

recall f(x)=y

y=(x+1)/(6x+7).....swap variables

x=(y+1)/(6y+7).....solve for y

x(6y+7)=(y+1)

6xy+7x=y+1

6xy-y=1-7x

(6x-1)y=1-7x

y=(1-7x)/(6x-1)

=> f '(x)=(1-7x)/(6x-1)

domain:
since denominator cannot be equal to zero, eliminate x value that makes it zero from the domain
6x-1=0 if x=1/6
so domain is: {x element R : x!=1/6} (where != means not equal)

range:
to find the range use quotient of highest degree variables which is
-7x/6x=>-7/6
so, range is: { f '(x)element R : f '(x)!=-7/6} (where != means not equal)

 

[MathLover1]

given: f (x)=(x+1)/(6x+7)

find f'(x)

recall f(x)=y

y=(x+1)/(6x+7).....swap variables

x=(y+1)/(6y+7).....solve for y

x(6y+7)=(y+1)

6xy+7x=y+1

6xy-y=1-7x

(6x-1)y=1-7x

y=(1-7x)/(6x-1)

=> f '(x)=(1-7x)/(6x-1)

domain:
since denominator cannot be equal to zero, eliminate x value that makes it zero from the domain
6x-1=0 if x=1/6
so domain is: {x element R : x!=1/6} (where != means not equal)

range:
to find the range use quotient of highest degree variables which is
-7x/6x=>-7/6
so, range is: { f '(x)element R : f '(x)!=-7/6} (where != means not equal)

 

[nycmathguy]

given: f (x)=(x+1)/(6x+7)

find f'(x)

recall f(x)=y

y=(x+1)/(6x+7).....swap variables

x=(y+1)/(6y+7).....solve for y

x(6y+7)=(y+1)

6xy+7x=y+1

6xy-y=1-7x

(6x-1)y=1-7x

y=(1-7x)/(6x-1)

=> f '(x)=(1-7x)/(6x-1)

domain:
since denominator cannot be equal to zero, eliminate x value that makes it zero from the domain
6x-1=0 if x=1/6
so domain is: {x element R : x!=1/6} (where != means not equal)

range:
to find the range use quotient of highest degree variables which is
-7x/6x=>-7/6
so, range is: { f '(x)element R : f '(x)!=-7/6} (where != means not equal)

Nicely done. This is algebra 2 math for sure. I agree about the tutors. We need more people. I would be more than happy to distribute cards in the NYC to students seeking help.