Rolle's Theorem - # 2

Jason76

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Verify Rolle's Theorem and then check for values of x\displaystyle x that satisfy it.

f(x)=x19x\displaystyle f(x) = \sqrt{x} - \dfrac{1}{9}x for interval [0,81]\displaystyle [0,81]

f(x)=(x)1/219x\displaystyle f(x) = (x)^{1/2} - \dfrac{1}{9}x

1st check: The function is differentiable and continuous over the interval, cause it's a polynomial.

f(0)=(0)1/219(0)=0\displaystyle f(0) = (0)^{1/2} - \dfrac{1}{9}(0) = 0

f(81)=(81)1/219(81)=0\displaystyle f(81) = (81)^{1/2} - \dfrac{1}{9}(81) = 0

2nd check

f(0)=f(81)\displaystyle f(0) = f(81)

f(x)=12u1/219\displaystyle f'(x) = \dfrac{1}{2}u^{-1/2} - \dfrac{1}{9}

f(x)=12x1/219\displaystyle f'(x) = \dfrac{1}{2}x^{-1/2} - \dfrac{1}{9}

12x1/219=0\displaystyle \dfrac{1}{2}x^{-1/2} - \dfrac{1}{9} = 0

12x1/2=19\displaystyle \dfrac{1}{2}x^{-1/2} = \dfrac{1}{9}

x1/2=29\displaystyle x^{-1/2} = \dfrac{2}{9}

[x1/2]2=[29]2\displaystyle [x^{-1/2}]^{-2} = [\dfrac{2}{9}]^{-2}

x=\displaystyle x = :?:
 
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Verify Rolle's Theorem and then check for values of x\displaystyle x that satisfy it.

f(x)=x19x\displaystyle f(x) = \sqrt{x} - \dfrac{1}{9}x for interval [0,81]\displaystyle [0,81]

f(x)=(x)1/219x\displaystyle f(x) = (x)^{1/2} - \dfrac{1}{9}x

1st check: The function is differentiable and continuous over the interval, cause it's a polynomial.
Yes, the function is diffeferentiable and continuous but it is not a polynomial.

f(0)=(0)1/219(0)=0\displaystyle f(0) = (0)^{1/2} - \dfrac{1}{9}(0) = 0

f(81)=(81)1/219(81)=0\displaystyle f(81) = (81)^{1/2} - \dfrac{1}{9}(81) = 0

2nd check

f(0)=f(81)\displaystyle f(0) = f(81)

f(x)=12u1/219\displaystyle f'(x) = \dfrac{1}{2}u^{-1/2} - \dfrac{1}{9}
Where did this "u" come from?

f(x)=12x1/219\displaystyle f'(x) = \dfrac{1}{2}x^{-1/2} - \dfrac{1}{9}

12x1/219=0\displaystyle \dfrac{1}{2}x^{-1/2} - \dfrac{1}{9} = 0

12x1/2=19\displaystyle \dfrac{1}{2}x^{-1/2} = \dfrac{1}{9}

x1/2=29\displaystyle x^{-1/2} = \dfrac{2}{9}

[x1/2]2=[29]2\displaystyle [x^{-1/2}]^{-2} = [\dfrac{2}{9}]^{-2}

x=\displaystyle x = :?:
So, after all this you can't do the arithmetic?

What is 22\displaystyle 2^2? What is 92\displaystyle 9^2? What does the negative power do?
 
Yes, the function is diffeferentiable and continuous but it is not a polynomial.


Where did this "u" come from?


So, after all this you can't do the arithmetic?

What is 22\displaystyle 2^2? What is 92\displaystyle 9^2? What does the negative power do?

I would come up with x=1(29)2\displaystyle x = \dfrac{1}{(\dfrac{2}{9})^{2}} which equals 1481\displaystyle \dfrac{1}{\dfrac{4}{81}}
 
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I would come up with x=1(29)2\displaystyle x = \dfrac{1}{(\dfrac{2}{9})^{2}} which equals 1481\displaystyle \dfrac{1}{\dfrac{4}{81}}
And? What is a much simpler way of writing 1481\displaystyle \dfrac{1}{\dfrac{4}{81}}?
 
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