1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Rolles:
    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!
    Lucas
     
  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)
    4sin(t)=1
    sin(t) = 1/4

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

    3) I think you dropped a minus sign, I get:
    2=sin(x/2)
    yeah no real min/max solutions

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

    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.

    Thanks
     
  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.

    f(x)=x^3-15x+c
    f(d)=d^3-15d+c
    0=d^3-15d+c
    -c=d^3-15d
    f(-2)=22=-c
    c=-22 when for f(-2)
    f(2)=-22=-c
    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.

    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Absolute min/max (algebra+solving for zero), Rolle's Thrm
  1. Absolute Max/Min (Replies: 2)

  2. Absolute Max Min Problem (Replies: 13)

  3. Absolute max/min (Replies: 12)

Loading...