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anonymous

  • one year ago

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval: x^4+x-3=0, (1,2)

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  1. anonymous
    • one year ago
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    Ok the interval is (1,2) so they are looking for you to use the Int Value Thm.... Do you know what this theorem means?

  2. anonymous
    • one year ago
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    Try and explain it to me and I will see if you nail it.... formulas are important but understanding is more so

  3. anonymous
    • one year ago
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    Ok well my teacher gave me the theorem and said if f is continuous on an interval and N is any number between f(a) and f(b) then there exists C on (a,b) such that f(c)=N. Not sure I understand that though....

  4. anonymous
    • one year ago
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    Ok let discuss then.... the easiest way is to think in terms of what happens when you use it to find a zero of a polynomial... like your problem for instance

  5. anonymous
    • one year ago
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    Since f is continuous there are no holes, jumps, gaps, kinks, etc.... so it can be represented by a smooth curve

  6. anonymous
    • one year ago
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    now lets take the interval from (a,b) and stipulate that f(a) is positive and f(b) is negative (i.e. f(a)>0 and f(b)<0)

  7. anonymous
    • one year ago
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    Since f is continuous the graph of the function must have crossed the x-axis somewhere inside the interval in order for the value of the function to change signs

  8. anonymous
    • one year ago
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    |dw:1442891184392:dw|

  9. anonymous
    • one year ago
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    Sorry its kind of messy, but this picture illustrates this point

  10. anonymous
    • one year ago
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    Now this relies specifically on the fact the end points are positive and negative... so we KNOW that the function must have crossed the x-axis AT LEAST once (though it could in principle cross way more than once)

  11. anonymous
    • one year ago
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    Does this make sense?

  12. anonymous
    • one year ago
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    Yes thank you...I get all of that. What I don't understand now is how this applies to the problem

  13. anonymous
    • one year ago
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    Oh well that is why I chose the example I did.... in that definition you stated the same thing is going on only I just chose to pick the value of N to be zero

  14. anonymous
    • one year ago
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    So the goal of the question is for you to evaluate the given function with the two end points of the interval

  15. anonymous
    • one year ago
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    You should find that one side is negative and one side is positive.... Since the function is continuous on that interval... the only way the function could change signs (from positive to negative) is if it crosses the x-axis.... i.e. has a zero

  16. anonymous
    • one year ago
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    in this case c would correspond to the (one of but not all!!) zeros and hence f(c)=N=0

  17. anonymous
    • one year ago
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    correspond to the zero* (singular)

  18. anonymous
    • one year ago
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    How am I supposed to get that zero? Plugging in 1 to the equation?

  19. anonymous
    • one year ago
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    This is a fourth order polynomial so it could have up to 4 real roots... but this mean value trick with this specified interval will only give us 1... we could repeat the process, but with only one interval we only have 1 shot to identify potentially 1 zero

  20. anonymous
    • one year ago
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    No the equation (aka the function f(x)) isnt supposed to equal zero.... the interval you were given (1,2) is specified by its endpoints 1 and 2

  21. anonymous
    • one year ago
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    evaluate f(1) and f(2) and show that one of them is positive and one of them is negative

  22. anonymous
    • one year ago
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    the only way this can happen AND f(x) be a continuous function is if f crosses the x-axis at some point (we dont know where) inside of this interval

  23. anonymous
    • one year ago
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    this point that it crosses the x-axis will correspond to the c you had in your above definition and the "y-value" (aka the value of the function) N will equal zero (we define it to be that since we are asked to look for a zero

  24. anonymous
    • one year ago
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    Does this make sense?

  25. anonymous
    • one year ago
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    Yes thank you so much for your patience and help!

  26. anonymous
    • one year ago
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    No sweat dude :D

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