inverse, hyperbolic, 1-1 functions; log equations; etc

FFEMT

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Nov 30, 2006
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I do not understand these 4 questions! I have looked in my book, and can't figure them out.

1) Find the inverse of 2.5y = x^2 - 6.2x + 12.61.

It says to graph, but I can't figure out the problem.

2) Find the hyperbolic sine and cosine of: sinh (x)=1/2 (e to the x power - e to the negative x power) and cosh (x)=1/2 (e to the x power + e to the negative x power) Show that sinh (x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)

3) Determine if these equations are one-to-one functions:
a) f(x) = 4x + 5/3x - 8
b) g(x) = 1/x^2- 2x + 3

4) Find all x which satisfy log base 2 x = log base 4 log base 2 3

** I have been looking and looking and cant figure them out. I really do not understand math and could use some help. THanks to all in advance!! **
 
So your class didn't cover any of these topics, and they're nowhere in your book, and you never saw any of this in any of your previous courses...?

Since the topics you reference entail months of classroom study, it would appear that you have been placed incorrectly. It would take a few weeks of intensive one-on-one study to cover this material, which obviously we cannot do within this environment.

It might be wise to conference with your academic advisor regarding being placed in a more appropriate course. My best wishes to you.

Eliz.
 
There is examples, but they are different. My instructor makes his own up, and tried to trick calculaters. So they are a little difficult to find in the book.
 
I do not need the answers, I would just like a place to start, so I can do it on my own!
 
FFEMT said:
3) Determine if these equations are one-to-one functions:
a) f(x) = 4x + 5/3x - 8
b) g(x) = 1/x^2- 2x + 3

A function is 1-1 if different numbers give different answers.

You've probably seen the mathematical definitions and had a rough time understanding what they meant?. Right?.

"A function f with domain D and range R is one-to-one if:

1.Whenever \(\displaystyle a\neq{b}\) in D, then \(\displaystyle f(a)\neq{f(b)}\) in R.

2. Whenever f(a)=f(b) in R, then a=b in D."


Put simply, when you put something in you get something unique out.

For instance, \(\displaystyle x^{2}\) is NOT one-to-one because \(\displaystyle (-1)^{2}\;\ and \;\ (1)^{2}\) = 1.

See?. in order for a function to be one-to-one we need to get different answers from different values. Here we got the same answer from 2 different values.

Also, for the inverse of a function to be defined it is essential that different numbers in the domain give different values of f.

Let's check and see whether your function is one-to-one or not.

\(\displaystyle \L\\f(x)=\frac{4x+5}{3x-8}\)

As mentioned at the top, suppose f(a)=f(b) in the domain of f.

\(\displaystyle \L\\\frac{4a+5}{3a-8}=\frac{4b+5}{3a-8}\)

\(\displaystyle \L\\(4a+5)(3b-8)=(4b+5)(3a-8)\)

\(\displaystyle \L\\12ab-32a+15b-40=12ab-32b+15a-40\)

\(\displaystyle \L\\-32a+15b=-32b+15a\)

\(\displaystyle \L\\a=b\)

See that?. That's pretty much what #2 in the definition at the top means.

Also, if you find your inverse and sub it in you get back to what they call the 'identity'.

\(\displaystyle \L\\f(x)=\frac{4x+5}{3x-8}\)

Solve for x in terms of y:

\(\displaystyle \L\\x=\frac{8y+5}{3y-4}=f^{-1}(y)\)

Now, we'll verify some things:

If should be able to sub \(\displaystyle f^{-1}(x)\) into f(x) and get x, \(\displaystyle f(f^{-1}(x))\)

\(\displaystyle \L\\\frac{4\left(\frac{8x+5}{3x-4}\right)+5}{3\left(\frac{8x+5}{3x-4}\right)-8}=x\)

Yep.

Now, try \(\displaystyle f^{-1}(f(x))=x\) for every x in the domain of \(\displaystyle f(x)\)

We sub \(\displaystyle f(x)\) into \(\displaystyle f^{-1}(x)\) and see if we get x.

\(\displaystyle \L\\f^{-1}(f(x))=x\) for every x in the domain of f.

\(\displaystyle \L\\\frac{8\left(\frac{4x+5}{3x-8}\right)+5}{3\left(\frac{4x+5}{3x-8}\right)-4}=x\)

Yep, that checks, too.

We are definitely one-to-one.

That was probably overkill, but I hope you have a better understanding now of what one-to-one is. Another name is injection

For your other function to check. If you can find two numbers that give you the same answer then it's NOT one-to-one.

I believe that x^2 is a hint that it's not, but you find out.

.
 
\(\displaystyle \L \begin{array}{rcl}
\sinh (x) & = & \frac{1}{2} \\
\frac{{e^x - e^{ - x} }}{2} & = & \frac{1}{2} \\
e^x - e^{ - x} & = & 1 \\
e^{2x} - e^x - 1 & = & 0 \\
\end{array}.\)

Now solve the last equation.
 
sorry, I have been workin, I will get back to you and tell you the answers! Thanks to those who are helping, it helps alot!
 
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