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Math Help - functions and models

  1. #1
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    functions and models

    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).
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nuttynutz View Post
    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
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nuttynutz View Post
    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
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    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
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nuttynutz View Post
    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
    Attached Thumbnails Attached Thumbnails functions and models-log.jpg  
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nuttynutz View Post

    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
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nuttynutz View Post
    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
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nuttynutz View Post
    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
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  9. #9
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    Quote Originally Posted by topsquark View Post
    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...
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    Quote Originally Posted by topsquark View Post
    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?
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    Quote Originally Posted by topsquark View Post
    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
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    Quote Originally Posted by topsquark View Post
    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
    Quote Originally Posted by nuttynutz View Post
    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
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    Quote Originally Posted by topsquark View Post
    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
    Quote Originally Posted by nuttynutz View Post
    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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