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Calculus problem.

  1. Oct 30, 2004 #1
    first problem : Let A and B be postitive numbers. SHow that not both of the numbers a(1-b) and b(1-a) can be greater than 1/4.

    second problem : Find a function f such that fprime(-1) = 1/2,fprime(0) = 0
    and f doubleprime (x) >0 for all x, or prove that sucha function cannot exist.

    thanks in advance!
  2. jcsd
  3. Oct 30, 2004 #2


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    plotting this on the ab plane gives the boundary of the regions where a(1-b)<1/4 and where a(1-b)>1/4. Starting with an (a,b) on the curve, adding a little to a will obviously give a(1-b)>1/4, and subtracting a little from a will give a(1-b)<1/4, so the region we want is above the curve. swapping a and b to get the second inequality is the same as reflecting this region over the line a=b, so to show these regions do not overlap, all you have to do is show the hyperbola does not cross the line a=b.

    the second question is easier. if f''(x) is always positive, f'(x) is always increasing as you move left to right.
    Last edited: Oct 30, 2004
  4. Oct 30, 2004 #3


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    For the second one.If f''(x)>0 for all x, then the graph of the function 'bends upwards', it is convex. That means f' is always increasing.
    (I assumed f''(x)>0 for all x, implies f''(x) exists for anyl real number x).
  5. Oct 30, 2004 #4
    thanks for the quick response guys!
  6. Oct 30, 2004 #5
    so showing that a= 1/(4-4b) a= (4b-1)/4b gotten from a(1-b) =1/4, b(1-a)=1/4
    only intercept once shows that there is no where that this is true
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