Rolle theroem, Mean Value Theroem

[abby_07]

I need some help on how to do Rolle's Theroem
f(x)=4x-tan(pi x) [-1/4,1/4]

these are the instructions: determine whether Rolle's Theroem can be applied to f on the indicated interval. Then if it can be applied find all the values of c in the interval such that f(c)=0

this is what i have done so far
f(-1/4)=0
f(1/4)=0
f'(x)= 4(cos pi x)^2- pi/(cos pi x)^2

4cos pi x^2- pi=0
i dont know what to do after that


Also i have no idea how to start Mean Value Theroem

f(x)=x(x^2-x-2) [-1,1]

f(x)=2sinx+sin2x [0, pi]

i am sorry this is a lot but if you tell me the steps i can probably figue out how to do the problems to the Mean Value Theroem

thanks

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[galactus]

Also i have no idea how to start Mean Value Theroem

\(\displaystyle f(x)=x(x^{2}-x-2), [-1,1]\)

Here's one. Try the next one yourself. If you get stuck, write back with your progress.

\(\displaystyle f(x)=x^{3}-x^{2}-2x\)

Because f is a polynomial, f is continuous and differentiable everywhere.

Hence, therefore, and ergo, it is continuous and differentiable on [-1,1].

So, the MVT is satisfied with a=-1 and b=1.

\(\displaystyle f(a)=f(-1)=0\)

\(\displaystyle f(b)=f(1)=-2\)

\(\displaystyle f'(x)=3x^{2}-2x-2\)

\(\displaystyle f'(c)=3c^{2}-2c-2\)

So, the equation:

\(\displaystyle \L\\f'(c)=\frac{f(b)-f(a)}{b-a}\)

becomes:

\(\displaystyle \L\\3c^{2}-2c-2=\frac{0-(-2)}{-2-0}=-1\)

Solve for c and we find \(\displaystyle \L\\c=1\;\ and \;\ \frac{-1}{3}\)

So c=1 and -1/3 are the numbers whose existence is guaranteed by the MVT.

Incidentally, the two lines are:

\(\displaystyle y=-x-1\) and \(\displaystyle {-}x+\frac{5}{27}\)

 

[abby_07]

WHERE DID THE 0-(-2)/-2-0 COME FROM

 

[stapel]

The Mean Value Theorem.

Eliz.

 

[abby_07]

i hope this is right , although i am stuck on how to find c
f(x)=2sinx+sin2x
f(0)=0
f(pi)=0
f'(x)=2cos2x+2cosx
2cos2x+2cosx=0-0/0-pi=0
how do i solve for c

 

[skeeter]

2cos(2x) + 2cos(x) = 0

cos(2x) + cosx = 0

use the identity cos(2x) = 2cos²(x) - 1 ...

2cos²(x) + cos(x) - 1 = 0

[2cos(x) - 1][cos(x) + 1] = 0

can you finish from here?