Why Might Graphs and Textbook Answers Differ on Function Ranges?

[Jacobpm64]

I have a few problems that i need help with...

Find the inverse of the function and verify that it is the inverse by performing a composition of functions both ways...

1. f(x) = (2x + 1) / (x + 3)

when i interchange x and y.. i can't seem to solve for y... because i have a y in the numerator and denominator...

Draw the graph and determine the domain and range of the function.

2. y = 2 ln (3-x) - 4

when i graph this equation on a calculator.. it doesn't look like the range is all reals... but in the back of the book.. the answer says the range is "all reals"... why?

3. y = log₂(x+1)

same problem with range here.. doesn't look like all reals.. but it is...

Solve each equation algebraically.

4. e^x + e^-x = 3

I can't seem to work this out anyway i try.. the x's end up canceling out.. and i lose my variable altogether

5. 2^x + 2^-x = 5

same problem here as the previous question

and lastly...

Solve for y.

6. ln (y - 1) - ln 2 = x + ln x

I don't know where to go with that one at all...

Thanks a lot in advance!

 

[Fermat]

1. x = (2y+1)/(y+3)

now cross-multiply and solve for y

2. Domain is for x-values, range is for y-values
As x gets very -ve (3 - x) gets very +ve and y gets very +ve. As x tends to -infinity, y tends to +infinity.
As x tends towards 3, (3-x) tends towards zero and lnx (and hence y) tends towards - infinity.

3. similar solution as above.

4. substitute for u = e^x and see if that helps.

5. similar solution as above.

6. use x = x.ln(e) or, x = ln(e^x)

 

[Jacobpm64]

ok.. this is where i get on the first one..

x = (2y+1)/(y+3)

x(y+3) = 2y + 1

Then i can divide by x, but i don't think that helps me... I'm stuck there


I have no idea what you're talking about in number 2 when you said -ve and +ve


and i don't know what you mean when you say "substitute for u = e^x and see if that helps." in #4.

Thanks for the help...

 

[roger]

you just expand the brackets and then put y outside the brackets .


he means to solve the quadratic equation you get by making the subtitution of e^x=u

so that u^2+3u+1=0

+ve is positive and -ve is negatibe

 

[HallsofIvy]

You have a serious problem. If you can't solve an equation like x(y+3) = 2y + 1 for y then you are going to have major problems with logarithms. I recommend you go to your teacher and ask for suggestions to help you with basic algebra.

 

[Brad Barker]

4. substitute for u = e^x and see if that helps.

5. similar solution as above.

yeah, and recall that e^(-x) is the same as 1/e^x



(my first hunch was to use hyperbolic cosine! )

 

[Fermat]

...
(my first hunch was to use hyperbolic cosine! )

Good idea. Faster than my substitution.

And it has a neater solution.

 

[jai6638]

4. substitute for u = e^x and see if that helps.

Hmmm. it doesn't seem to factor nicely. Am I doing something wrong here?

hyperbolic cosine!

whats that?

 

[Brad Barker]

Hmmm. it doesn't seem to factor nicely. Am I doing something wrong here?

yeah, probably. you should get a quadratic equation where you can solve for u using the quadratic formula.

in fact, it's ALMOST that quadratic equation roger put in his post... except one of his signs is wrong.

whats that?

hyperbolic cosine is written as "cosh."

cosh x := (e^x + e^(-x))/2

(and, like the standard cosine, hyperbolic cosine has an inverse function with the usual property that the composition of the function and its inverse will get you back "x.")

but... you won't need it to solve this problem, it turns out. the hyperbolic functions are pretty obscure--you typically won't even deal with them in the calc sequence!

 

[jai6638]

yeah, probably. you should get a quadratic equation where you can solve for u using the quadratic formula.

in fact, it's ALMOST that quadratic equation roger put in his post... except one of his signs is wrong.

I did get all that... However, if I use the quadratic formula, the solutoin I get does not equal to three which is what I think it is suppose to equal as it states in the question...

 

[VietDao29]

I did get all that... However, if I use the quadratic formula, the solutoin I get does not equal to three which is what I think it is suppose to equal as it states in the question...

Why must the solution equal to 3? The solution x can be anything that satisfy:
e ^ x + e ^ {-x} = 3
When you have found out x, just plug x in the equation and see if it returns 3. If it does, then you are correct.
Just remember when you let u = e ^ {x}
And you have worked out the u that satisfied u ^ 2 - 3u + 1 = 0, you have to change the u back to x, ie:
u = e ^ {x} \Leftrightarrow x = \ln u
Viet Dao,

 

[jai6638]

ahh I see. I don't know why I thought it must equal 3... Thanks much