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

1. Nov 23, 2008

### LucasCLarson

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. Nov 24, 2008

### LucasCLarson

anyone?

3. Nov 24, 2008

### LogicalTime

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

4. Nov 24, 2008

### LogicalTime

1)
$f(x) = (x^{ \frac{1}{3} })(4-x^2)^{\frac{1}{2}}$
$f'(x) = ( \frac{1}{3}x^{ - \frac{2}{3} })(4-x^2)^{\frac{1}{2}} +(x^{ \frac{1}{3} })( \frac{1}{2})(4-x^2)^{- \frac{1}{2}}(-2x)$
$0= ( \frac{1}{3}x^{ - \frac{2}{3} })(4-x^2)^{\frac{1}{2}} +(-x^{ \frac{4}{3} })(4-x^2)^{- \frac{1}{2}}$

give it a shot from here?

5. Nov 24, 2008

### LogicalTime

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

6. Nov 25, 2008

### LucasCLarson

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

7. Nov 25, 2008

### LucasCLarson

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