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anonymous

  • one year ago

Suppose f(x) = 2x - 1 and g(x) = x^2. Find (f+g)(x) and state domain.

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  1. jim_thompson5910
    • one year ago
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    (f+g)(x) is the same as f(x) + g(x) so you just add up the functions and combine like terms if possible

  2. anonymous
    • one year ago
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    x^2 + 2x - 1 I get that part, it's more so that I have trouble understanding how to find domain.

  3. jim_thompson5910
    • one year ago
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    Rule: the domain of ANY polynomial is the set of all real numbers. You can plug in any real number in for x, for any polynomial, and you'll get some real number out.

  4. jim_thompson5910
    • one year ago
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    There are no restrictions (eg: division by zero errors) to worry about

  5. anonymous
    • one year ago
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    Okay! Thank you. Can you help me w/ another problem?

  6. jim_thompson5910
    • one year ago
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    Sure, go ahead

  7. anonymous
    • one year ago
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    Find h(k(x)) and k(h(x)), and state the domain. h(x) = x^2 - 3x + 1 and k(x) = log5(x). So, \[(\log_{5} )^{2} - 3(\log_{5} ) + 1\]

  8. jim_thompson5910
    • one year ago
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    I think you meant to say \[\Large h(k(x)) = \left(\log_{5}(x)\right)^2 + 3\log_{5}(x) + 1\] right?

  9. anonymous
    • one year ago
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    Yeah, I forgot the x's.

  10. jim_thompson5910
    • one year ago
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    as for the domain of h(k(x)), any ideas?

  11. anonymous
    • one year ago
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    \[[1, \infty)\]

  12. jim_thompson5910
    • one year ago
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    close

  13. jim_thompson5910
    • one year ago
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    you can plug in numbers for x smaller than 1 like x = 0.5 but you cannot plug in x = 0 or anything negative

  14. jim_thompson5910
    • one year ago
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    so the domain is x > 0 which in interval notation is \(\Large (0,\infty)\)

  15. anonymous
    • one year ago
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    Okay, that makes sense. Thanks!

  16. jim_thompson5910
    • one year ago
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    how about k(h(x)) ?

  17. anonymous
    • one year ago
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    Would log5(x^2 - 3x + 1)'s domain be (-inf, +inf)?

  18. jim_thompson5910
    • one year ago
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    no

  19. jim_thompson5910
    • one year ago
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    hint: look at the graph of y = x^2 - 3x + 1. Are there points on that parabola that have y coordinates that are either 0 or a negative number?

  20. anonymous
    • one year ago
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    Yes, (1, -1) and (2, -1).

  21. anonymous
    • one year ago
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    Well, looking at the table tells me that.

  22. jim_thompson5910
    • one year ago
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    so because of that, log(x^2 - 3x + 1) = log(-1) when x = 1 but log(-1) leads to a non-real number. So x = 1 is NOT part of the domain

  23. jim_thompson5910
    • one year ago
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    there are other x values that aren't part of the domain for this reason

  24. anonymous
    • one year ago
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    Would it be\[(-\infty, 1] \cup [1, 2] \cup [2, \infty)\]

  25. jim_thompson5910
    • one year ago
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    what are the roots of x^2 - 3x + 1 ?

  26. anonymous
    • one year ago
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    Are there even any? Doing the big x, what is there that multiplies to 1 and adds to three?

  27. jim_thompson5910
    • one year ago
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    no, so you'll need to use the quadratic formula

  28. jim_thompson5910
    • one year ago
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    x^2 - 3x + 1 does not factor the roots are decimal values

  29. anonymous
    • one year ago
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    5/2 and 1/2?

  30. jim_thompson5910
    • one year ago
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    I graphed x^2 - 3x + 1 with a graphing calculator https://www.desmos.com/calculator/fp80fcskll click on the x-intercepts and the coordinates will pop up. Tell me what values pop up

  31. anonymous
    • one year ago
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    0.382 2.618 I'm so mad b/c I got those before but thought they were wrong.

  32. jim_thompson5910
    • one year ago
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    notice how the portion between 0.382 and 2.618 is below the x axis the points on the parabola in this interval have negative y coordinates or the y coordinates are 0. So we have to kick this interval out of the domain Start with \(\Large (-\infty, \infty)\) and kick out that interval to end up with \(\Large (-\infty, 0.382)\cup(2.618, \infty)\)

  33. jim_thompson5910
    • one year ago
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    So the domain of log(x^2 - 3x + 1) is \(\Large (-\infty, 0.382)\cup(2.618, \infty)\)

  34. anonymous
    • one year ago
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    Thank you for all of your help. I think I understand what to do now.

  35. jim_thompson5910
    • one year ago
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    sure thing, glad to be of help

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