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Absolute min/max (algebra+solving for zero), Rolle's Thrm

  1. Nov 23, 2008 #1
    Hey everyone. Just getting prepped for a midterm on Tuesday and looking for a bit of help on a few things. If there are any tricks to make some of this stuff easier that would be great. I remember there being a few from back in high school, but i can't remember them. I know the process for all of these, but some of the algebra is pretty tough. Non of the answers are posted so it would be great to have a bit more insight.

    We aren't allowed to use calculators either.

    1. The problem statement, all variables and given/known data
    Solving these equations [f'(x)] for 0 to determine the critical points (min/max) of the original function:
    1) f(x)=(x^(1/3))((4-x^2)^(1/2)) [-1,2] ;
    f'(x)=(sqrt(4-x^2)/3((x^2)^(1/3)) + x^(1/2)(.5(4-x^2)^(-1/2)

    2) f(t)=sint+cos2t [0,pi/2]; f'(t)= -cost-2sin2t
    3) f(x)=x+cos(x/2) [pi/4,7pi/4]; f'(x)=1-(1/2)sin(x/2)
    4) f(x)=x-lnx [1/2,2]; f'(x)= 1-(1/x) (this seems easy, but when i try to determine the intervals of increase and decrease it makes no sense. the C.P. should be at x=1 [where f'(x)=0], but the intervals don't show a max or min.)

    Rolles: I can't grasp the concept of solving by contradiction.

    5)Show the equation 1+2x+x^3+4x^5=0

    3. The attempt at a solution

    1) i tried to find a common denominator but wasn't sure were to go. If you multiply two sqrt functions both obviously cancel each other out, but do you still multiply/foil the stuff inside the sqrt?? sqrt(4-x^2) x 2sqrt(4-x^2) ??

    2) difficult to grasp without a clac. Any pointers on solving mathematically. I know all of the intervals of pi, i also recognized that if i can find the intersection of cos & sin of the original function on the interval i should have my answer.

    3) No idea: 0=1-.5sin(x/2) = -2=sin(x/2) (function never reaches -2)

    4)Interval problem:

    5)Use IVT to show that a root exits. I chose a=-1 and b=1 f(a)<0<f(b), and because f is continuous(polynomial) a root exists for f(x) on the interval (-1,1) such that f(c)=0
    F is a polynomial so it is certainly continuous and differentiable.
    and f(a)=f(b) then there must be a number N such that f'(c)=0

    My problem: Am i supposed to use the interval i used to prove there was a root ([-1,1]) or doesn't it matter? My book shows that f(a)=0=f(b); is that because it was the root given for my original equation or because i know there must be a point such that f'(c)=0. I know that f'(x)=2+3x^2+20x^4 will always be greater than or equal to 2. Now, how to i prove that only one root exits by contradiction???

    Thank you very much!
  2. jcsd
  3. Nov 24, 2008 #2
  4. Nov 24, 2008 #3
    I think you're making some minus sign dropping errors!

    2) f(t) = sin(t)+cos(2t)
    f'(t) =cos(t)-2sin(2t)
    0 = cos(t)-2sin(2t)
    2sin(2t) = cos(t)
    2(sin(t)cos(t) +sin(t)cos(t)) =cos(t)
    4sin(t)cos(t) =cos(t)
    sin(t) = 1/4

    don't forget the solutions where cos(t) = 0

    3) I think you dropped a minus sign, I get:
    yeah no real min/max solutions

    4) graphing it on your calculator might help here
  5. Nov 24, 2008 #4
    f(x) = (x^{ \frac{1}{3} })(4-x^2)^{\frac{1}{2}}
    f'(x) = ( \frac{1}{3}x^{ - \frac{2}{3} })(4-x^2)^{\frac{1}{2}} +(x^{ \frac{1}{3} })( \frac{1}{2})(4-x^2)^{- \frac{1}{2}}(-2x)
    0= ( \frac{1}{3}x^{ - \frac{2}{3} })(4-x^2)^{\frac{1}{2}} +(-x^{ \frac{4}{3} })(4-x^2)^{- \frac{1}{2}}

    give it a shot from here?
  6. Nov 24, 2008 #5
    5) since the derivative is always positive the function is always increasing. It's also continuous therefore it can only pass the y=0 line (and every other line y = constant) once if at all.

    hope these help, think them through and then let me know if anything was confusing :cool:
  7. Nov 25, 2008 #6
    thanks a lot for the help. I was able to get them figured out! I was being an idiot on the 1-1/x fcn.

    I have another Rolles question as well. I will post it at the top.

  8. Nov 25, 2008 #7
    Nevermind, looks like i can't edit for some reason.

    here is my question.

    show x^3-15x+c=0 has at most one root on the interval [-2, 2].

    My work/attempted solution.

    Using the IVT i determined that there is a possible root (N=c) within the interval of [-2,2]. because c is my constant i decided to Use [d] as my letter for explanation.

    c=-22 when for f(-2)
    c=22 for f(2)

    f is continuous on [-2,2] because it is a polynomial. Let -c be a number b/w f(-2) and f(2) where f(-2) does not equal f(2). Then there exist a root d on (-2,2) such that F(d)=-c

    I am stuck there. I feel like i have done a bunch of stuff wrong so a little nudge in the right direction would be great.

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