zeesbrat3
  • zeesbrat3
If f(x) = |(x2 − 4)(x2 + 2)|, how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?
Mathematics
katieb
  • katieb
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zeesbrat3
  • zeesbrat3
zeesbrat3
  • zeesbrat3
help..
Michele_Laino
  • Michele_Laino
I'm pondering...

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zeesbrat3
  • zeesbrat3
Oh, sorry
Michele_Laino
  • Michele_Laino
we have to note that we have the subsequent inequality: \[\Large {x^2} - 4 < 0,\quad {\text{if }} - 2 < x < 2\]
Michele_Laino
  • Michele_Laino
am I right?
zeesbrat3
  • zeesbrat3
honestly, im confused
Michele_Laino
  • Michele_Laino
why?
Michele_Laino
  • Michele_Laino
it is a simple inequality
zeesbrat3
  • zeesbrat3
no i understand the inequality, im confused by how you got it
Michele_Laino
  • Michele_Laino
ok! I wrote that inequality, because inside the interval [0,1] your function can be rewritten as follows: \[\Large f\left( x \right) = - \left( {{x^2} - 4} \right)\left( {{x^2} + 2} \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)\]
zeesbrat3
  • zeesbrat3
otherwise any other would be out of bounds
Michele_Laino
  • Michele_Laino
no, it is a step in order to get rid the absolute value symbol
zeesbrat3
  • zeesbrat3
oh, okay.
Michele_Laino
  • Michele_Laino
now, we can apply the theorem af mean value
Michele_Laino
  • Michele_Laino
so, we have to compute this: \[\Large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = ...\]
zeesbrat3
  • zeesbrat3
\[f'(c) = \frac{ f(b) - f(a) }{ b - a }\]
zeesbrat3
  • zeesbrat3
1
Michele_Laino
  • Michele_Laino
are you sure?
Michele_Laino
  • Michele_Laino
\[\large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = \frac{{6 - 8}}{1} = ...?\]
zeesbrat3
  • zeesbrat3
isnt 3x3 9?
Michele_Laino
  • Michele_Laino
sorry I have made a big error, you are right
zeesbrat3
  • zeesbrat3
it's okay :)
Michele_Laino
  • Michele_Laino
sorry again!!
zeesbrat3
  • zeesbrat3
no apologies necessary! thank you for your help
Michele_Laino
  • Michele_Laino
next we have to compute the first derivative at a generic point x, namely we have to compute the first derivative of this function: \[\Large f\left( x \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)\]
Michele_Laino
  • Michele_Laino
what do you get?
zeesbrat3
  • zeesbrat3
\[-4x(x^2-1)\]
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
then we have to solve this equation: \[ \Large - 4x\left( {{x^2} - 1} \right) = 1\]
Michele_Laino
  • Michele_Laino
it is not a simple equation
zeesbrat3
  • zeesbrat3
\[x^3 = -\frac{ 1 }{ 2 }\] ?
Michele_Laino
  • Michele_Laino
how did you do to write that expression?
zeesbrat3
  • zeesbrat3
\[-4x(x^2-1) = 1\]\[\frac{ 1 }{ -4x } = x^2\] \[\frac{ 1 }{ -4x } +1 = x^2\] \[1+1 = -4x^3\] \[- \frac{ 1 }{ 2 } = x^3\]
Michele_Laino
  • Michele_Laino
I don't understand this step: \[\frac{1}{{ - 4x}} = {x^2}\]
Michele_Laino
  • Michele_Laino
anyway it is not a solution of our equation
zeesbrat3
  • zeesbrat3
Sorry, i forgot the -1
zeesbrat3
  • zeesbrat3
How would you get the proper solution?
Michele_Laino
  • Michele_Laino
I questioned wolfram alpha, here are the solutions:
1 Attachment
zeesbrat3
  • zeesbrat3
So, 3 solutions in the interval
Michele_Laino
  • Michele_Laino
we have two solutions, inside the interval [0,1]
zeesbrat3
  • zeesbrat3
Whoops, my bad! I looked outside the interval
Michele_Laino
  • Michele_Laino
1 Attachment
Michele_Laino
  • Michele_Laino
do you see them?
zeesbrat3
  • zeesbrat3
Yes. Since one of them is in the negatives, it is not in the interval and does not matter to answer the question
Michele_Laino
  • Michele_Laino
yes! that's right!
zeesbrat3
  • zeesbrat3
Thank you!
Michele_Laino
  • Michele_Laino
:)
zeesbrat3
  • zeesbrat3
I tried doing the same thing on this problem, but I keep getting no solutions: A particle moves along the x-axis with position function s(t) = xe^x. How many times in the interval [−5, 5] is the velocity equal to 0?
Loser66
  • Loser66
velocity = (s(t))' take derivative and let it =0 to solve for x .

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