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Homework Help: Calculus Problem

  1. May 23, 2005 #1
    Hi, this is my first post, and I don't want to seem too desparate but I really need help with a problem that I have been unable to do (I've been working on it for close to 2 hours).

    If function f(x) is continuous and differentiable and

    f(x) = ax*x*x – 6x when x is less than or equal to 1
    f(x) = bx*x + 4 when x is greater than 1

    then a = ?

    (ax*x*x meaning a * x-cubed, etc.)

    Anyways, I know that I'm not supposed to ask for a full solution, but I have been unable to do this problem for a very long time, and any help is appreciated (even a full solution).

    Thanks for any help.
  2. jcsd
  3. May 23, 2005 #2
    The function f(x) is a piecewise function, and it is continuous throughout. It is divided at f(1), but in order for it to be continuous it means that both pieces approach f(1) from both sides. Try to put what I just said together algebraically. If you can't I'll privde you with another hint.
  4. May 23, 2005 #3
    I'm sorry, but I don't understand what you mean.
  5. May 23, 2005 #4
    Continuous Function
    Piecewise Function

    Your funciton f(x) is continuous and piecewise.
  6. May 23, 2005 #5
    Well, I suppose that a continuous function would be one in which all x-coordinates have a corresponding y-coordinate?

    I'm not sure what a piecewise function is though. I'm guessing that it's one in which the function is defined differently in different ranges?
  7. May 23, 2005 #6
    @ whozum: I'm not really following your train of thought here, I'm not really very familiar with calculus-related terminology, so you might need to dumb it down a shake for me.

    If anyone else feels like jumping in too, feel free to do so.
  8. May 23, 2005 #7
    What you have is:

    [tex]f(x)=\left\{\begin{array}{cc}ax^3 - 6x,&\mbox{ if }
    x\leq 1\\bx^2 +4, & \mbox{ if } x>1\end{array}\right.[/tex]

    For continuity, yes, there would have to be a corresponding x and y coordinate. If there was a discontinuity then the function would have points where the function was undefined (jumps, holes, asymptotes....)

    For piecewise, again, pretty much so. A piecewise function is defined differently within different ranges.

    Going back to what Whozum's first post, now that you know that the function is continuous (no breaks in the function), but it is piecewise (defined differently at certain points) put the two together. You are given where the pieces of the function meet up when you are given where x is defined.

    Now take a look at the function at this point in two different ways...
  9. May 24, 2005 #8
    Since its continuous,

    [tex] \lim_{x\rightarrow 1^+} bx^2 + 4 = \lim_{x\rightarrow 1^-} ax^3-6x[/tex]
  10. May 24, 2005 #9


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    Homework Helper

    Condition for continuity at x = 1 :

    [tex]\lim_{x \rightarrow 1^-}f(x) = \lim_{x \rightarrow 1^+}f(x)[/tex]

    (as whozum already stated).

    Condition for the existence of first derivative (differentiability) at x = 1 :

    [tex]\lim_{x \rightarrow 1^-}f'(x) = \lim_{x \rightarrow 1^+}f'(x)[/tex]

    So what you have to do is to first get an equation in a and b by setting x = 1 in f(x), using both expressions. Equate the two expressions for f(x) at x = 1 to get one linear equation in a and b.

    Then differentiate the function to get f'(x) on either side of x = 1. Again set x = 1 in f'(x) and get another linear equation in a and b by equating the two expressions for f'(x) at x = 1.

    You now have a simple system of linear simultaneous equations in two variables. Solve for a and b, you have the answer.
  11. May 24, 2005 #10
    I suppose your problem is already solved. But I'll throw in a few lines nevertheless

    A piecewise function which is defined over different subintervals in different ways. The first example of a piecewise function that I can give you is the absolute value function, [itex]|x|[/itex]. By saying that something is a function, I do not mean that it's 'parts' are necessarily continuous at their respective end points. As a matter of fact, it is desirable in real life (as part of solutions to differential and algebraic equations which pertain to real life problems) that we have (sometimes) functions which are piecewise, to account for certain irregularities or changes in a process. As a result, their continuity at the end points is saying a very strong thing. And sometimes, as a result of experiments, parametrically defined piecewise functions are obtained, the values of parameters being decided by some additional constraints (this is where common sense takes over) which ensure continuity and possibly differentiability at the end points.

    So to sum up (and make sure this post doesn't become more boring than it is :tongue: already) in a problem such as this you are asked to remove a discontinuity. Discontinuities are of two kinds: removable and irremovable. If the left and right hand limits exist at a point but the function is either undefined or grows asymptotically about that point, then we call the discontinuity removable since we can remove it by a suitable redefinition of the function at that point (so the left, right limits and the function's value at that point all agree). However, if the limits do not exist in the first place or if they exist but are unequal then no matter how much we try, we can't remove the discontinuity by any redefinition of the function over the real set. For obvious reasons then, this is an irremovable discontinuity.

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