functions and models

**nuttynutz** · September 27th 2007, 07:30 AM

The following problems are the only problems from my homework that i couldn't not solve or rather figure out. If i could by any chance receive any help to that would be great. -please if you help me solve these show me the steps so i at least know how to do it... thank you.

1)Find an expression for the function whose graph consists of the line segment from the point (-2,2) to the point (-1,0) together with the top half of the circle with center the origin and radius 1.

F(x)= -2x-2 if -2 < x < -1
root(1-x^2) if -1< x <1

why?

2) if f(x) = ln x and g(x) = x^2-9 find the functions of a, b ,c ,d and their domains
a)(f o g) ln(x^2-9)
b)(g o f) (ln x)^2 -9
c) (f o f) ln(ln(x))
d) (g o g) (x^2-9)^2 -9 = x^4-18x^2 +72

3) if f(x) = 2x+ ln(x) find inverse of f(2)
the answer is 1 but how?

4) Solve each equation for x.
a) e^x = 5 x = 5th root of e? am i right?
b) ln(x) = 2
c)e^e^x = 2 x = ln(ln(2))

5) sketch the graph of the function g(x) = |x^2-1| - |x^2=4|

Any help with any of the above problems would be greatly appreciated it. Those are some of the few problems that i seem to be having a hard time. I also showed some of the work that i did in regular font(no bold).

**topsquark** · September 27th 2007, 08:26 AM

Originally Posted by nuttynutz

The following problems are the only problems from my homework that i couldn't not solve or rather figure out. If i could by any chance receive any help to that would be great. -please if you help me solve these show me the steps so i at least know how to do it... thank you.

1)Find an expression for the function whose graph consists of the line segment from the point (-2,2) to the point (-1,0) together with the top half of the circle with center the origin and radius 1.

F(x)= -2x-2 if -2 < x < -1
root(1-x^2) if -1< x <1

why?

The first part of the function is because $f(x) = -2x - 2$ is the line containing the points (-2, 2) and (-1, 0).

For the second part note that the circle of radius 1 and center at the origin is
$x^2 + y^2 = 1$

So solving for $y = f(x) = \sqrt{1 - x^2}$ .

-Dan

**topsquark** · September 27th 2007, 08:30 AM

Originally Posted by nuttynutz

2) if f(x) = ln x and g(x) = x^2-9 find the functions of a, b ,c ,d and their domains
a)(f o g) ln(x^2-9)
b)(g o f) (ln x)^2 -9
c) (f o f) ln(ln(x))
d) (g o g) (x^2-9)^2 -9 = x^4-18x^2 +72

So far so good. Now what the their domains?

-Dan

**nuttynutz** · September 27th 2007, 08:39 AM

i can't find their domain... i don't know how to solve for ln at least not yet. so, i can't find the domain

**topsquark** · September 27th 2007, 08:58 AM

Originally Posted by nuttynutz

3) if f(x) = 2x+ ln(x) find inverse of f(2)
the answer is 1 but how?

Don't you mean the inverse function evaluated at 2? ie. $f^{-1}(2)$ ?

Unless you get lucky and see the answer probably the best thing to do would be to graph the function and look for the x value such that f(x) = 2. See the graph below.

-Dan

**topsquark** · September 27th 2007, 09:01 AM

Originally Posted by nuttynutz

4) Solve each equation for x.
a) e^x = 5 x = 5th root of e? am i right?
b) ln(x) = 2
c)e^e^x = 2 x = ln(ln(2))

a)
$e^x = 5$

$ln(e^x) = ln(5)$

$x = ln(5)$

b)
$ln(x) = 2$

$e^{ln(x)} = e^2$

$x = e^2$

c)
$e^{e^x} = 2$

$ln \left ( e^{e^x} \right ) = ln(2)$

$e^x = ln(2)$

$ln(e^x) = ln(ln(2))$

$x = ln(ln(2))$

-Dan

**topsquark** · September 27th 2007, 09:13 AM

Originally Posted by nuttynutz

5) sketch the graph of the function g(x) = |x^2-1| - |x^2=4|

Worst comes to worst, pick a bunch of x values and find the g(x)'s.

$g(x) = |x^2 - 1| - |x^2 - 4|$

$g(x) = |(x + 1)(x - 1)| - |(x + 2)(x - 2)|$

Critical points are where the arguments of the absolute value bars is zero, so I have critical points at $x = -2, -1, 1, 2$ .

For $(-\infty, -2)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$ .

So on $(-\infty, -2)$ $g(x) = (x + 1)(x - 1) - (x + 2)(x - 2) = 3$

For $(-2, -1)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) < 0 \implies |(x + 2)(x - 2)| = -(x + 2)(x - 2)$ .

So on $(-2, -1)$ $g(x) = (x + 1)(x - 1) + (x + 2)(x - 2) = 2x^2 - 5$

So far we have:
$g(x) = \begin{cases} 3, \text{ for }-\infty \leq x \\ 2x^2 - 5, \text{ for } -2 < x \leq -1 \end{cases}$

I'll let you figure the other two intervals.

-Dan

**topsquark** · September 27th 2007, 09:15 AM

Originally Posted by nuttynutz

2) if f(x) = ln x and g(x) = x^2-9 find the functions of a, b ,c ,d and their domains
a)(f o g) ln(x^2-9)
b)(g o f) (ln x)^2 -9
c) (f o f) ln(ln(x))
d) (g o g) (x^2-9)^2 -9 = x^4-18x^2 +72

If you haven't worked with the ln function then why are you even trying to do these??

You should still be able to do d). This is simply a polynomial. What is the domain of a polynomial function?

-Dan

**nuttynutz** · September 27th 2007, 01:11 PM

Originally Posted by topsquark

If you haven't worked with the ln function then why are you even trying to do these??

You should still be able to do d). This is simply a polynomial. What is the domain of a polynomial function?

-Dan

all real numbers =) lol sometimes i don't think...

**nuttynutz** · September 27th 2007, 03:19 PM

Originally Posted by topsquark

Don't you mean the inverse function evaluated at 2? ie. $f^{-1}(2)$ ?

Unless you get lucky and see the answer probably the best thing to do would be to graph the function and look for the x value such that f(x) = 2. See the graph below.

-Dan

could this be done without the use of a calculator? if so then how?

**nuttynutz** · September 27th 2007, 05:19 PM

Originally Posted by topsquark

Worst comes to worst, pick a bunch of x values and find the g(x)'s.

$g(x) = |x^2 - 1| - |x^2 - 4|$

$g(x) = |(x + 1)(x - 1)| - |(x + 2)(x - 2)|$

Critical points are where the arguments of the absolute value bars is zero, so I have critical points at $x = -2, -1, 1, 2$ .

For $(-\infty, -2)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$ .

So on $(-\infty, -2)$ $g(x) = (x + 1)(x - 1) - (x + 2)(x - 2) = 3$

For $(-2, -1)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) < 0 \implies |(x + 2)(x - 2)| = -(x + 2)(x - 2)$ .

So on $(-2, -1)$ $g(x) = (x + 1)(x - 1) + (x + 2)(x - 2) = 2x^2 - 5$

So far we have:
$g(x) = \begin{cases} 3, \text{ for }-\infty \leq x \\ 2x^2 - 5, \text{ for } -2 < x \leq -1 \end{cases}$

I'll let you figure the other two intervals.

-Dan

alright i did a lot of research and sort of reviewing to try and figure everything out and i think i have it all done correctly my only problem is the question above.. i really am clueless when it comes to it. Any help would be appreciated it

**topsquark** · September 28th 2007, 05:14 AM

Originally Posted by topsquark

Don't you mean the inverse function evaluated at 2? ie. $f^{-1}(2)$ ?

Unless you get lucky and see the answer probably the best thing to do would be to graph the function and look for the x value such that f(x) = 2. See the graph below.

-Dan

Originally Posted by nuttynutz

could this be done without the use of a calculator? if so then how?

The only way to do it without a calculator is to look at the equation and see what your intuition says. I can't guide you with that any further, except to say that I would likely have (if I hadn't seen the solution right below it) tried a few integer x to see what came out of it. If you did that you would have noted that $f(1) = 2$ and thus you could easily get $f^{-1}(2) = 1$ .

-Dan

**topsquark** · September 28th 2007, 05:25 AM

Originally Posted by topsquark

Worst comes to worst, pick a bunch of x values and find the g(x)'s.

$g(x) = |x^2 - 1| - |x^2 - 4|$

$g(x) = |(x + 1)(x - 1)| - |(x + 2)(x - 2)|$

Critical points are where the arguments of the absolute value bars is zero, so I have critical points at $x = -2, -1, 1, 2$ .

For $(-\infty, -2)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$ .

So on $(-\infty, -2)$ $g(x) = (x + 1)(x - 1) - (x + 2)(x - 2) = 3$

For $(-2, -1)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) < 0 \implies |(x + 2)(x - 2)| = -(x + 2)(x - 2)$ .

So on $(-2, -1)$ $g(x) = (x + 1)(x - 1) + (x + 2)(x - 2) = 2x^2 - 5$

So far we have:
$g(x) = \begin{cases} 3, \text{ for }-\infty \leq x \\ 2x^2 - 5, \text{ for } -2 < x \leq -1 \end{cases}$

I'll let you figure the other two intervals.

-Dan

Originally Posted by nuttynutz

alright i did a lot of research and sort of reviewing to try and figure everything out and i think i have it all done correctly my only problem is the question above.. i really am clueless when it comes to it. Any help would be appreciated it

The solution method I gave you is pretty systematic and exploits the fact that
$|x| = \begin{cases} -x; x < 0 \\ x; x \geq 0 \end{cases}$

You look at your function and see where the argument inside the absolute value bars is negative and then equate that to the negative of the argument. Case in point:

For $(-\infty, -2)$ , $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$

from above. (Sorry, I guess I could've used some separators in there for clarity.)

I've given you examples of how to do it with two of the intervals, and I'm asking you to do it yourself for the other three intervals: $(-1, 1) \text{ and } (1, 2) \text{ and } (2, \infty )$ .

Just find where each expression inside the absolute value bars is negative, then replace the absolute value with the negative of that expression. Then simplify.

Edit: If the problem is notation, let me break down the above quote a bit:

For $(-\infty, -2)$ , $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$

This means that I am considering the case where x is between $-\infty$ and -2. So we know that $x + 1$ is negative and $x - 1$ is negative so $(x + 1)(x - 1)$ is positive, etc.

-Dan

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