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Homework Help: Problems getting started on this problem

  1. Sep 17, 2006 #1
    Hi I am trying to solve the following problem.

    f(x) = 2x^3 - 5x - 5
    show that there is atleast one value of c for which f(x) = pie

    I am assuming they want me to use pie = 3.14

    so I did the following

    2x^3 - 5x - 5 = 3.14

    but then I dont know how to solve for x can someone please give me some kinda hint to get me started?

    Thanks
     
  2. jcsd
  3. Sep 17, 2006 #2

    shmoe

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    I sure hope they don't!


    f(x) = 2x^3 - 5x - 5 is a continuous function (right?), what kinds of theorems or results do you have on continuous functions that might be useful here?
     
  4. Sep 17, 2006 #3
    ok so i have to use the intermediate value theorem. i still dont understand the pie thing... if its not 3.14 then what do they want me to use for pie.

    so with the IVT i would do the following

    1] define an inteval [a, b] such that when i find f(a) and f(b) one value is bigger than "pie" and the other value is less than "pie"

    2] I would also have to show that f is continous in that interval [a, b]

    so for example if i find out that f(a) < "pie" < f(b) then i can say that yes there does exist a point c where f(c) = "pie"

    but I dont know what "pie" is i guess so I'm not sure what I am looking for.
     
  5. Sep 17, 2006 #4

    shmoe

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    pi=3.141526535... it does not equal any truncated sum with a finite number of digits (it also has no "e" on the end). For your purposes you can probably safely assume some bounds like 3<=pi<=4 and find an a and b where:

    f(a) < 3 <= pi <= 4 < f(b)

    as for continuity, f(x) is a polynomial. You should know something about polynomials and continuity.
     
  6. Sep 18, 2006 #5
    thanks got it
     
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