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zeesbrat3
 one year ago
If f(x) = (x2 − 4)(x2 + 2), how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?
zeesbrat3
 one year ago
If f(x) = (x2 − 4)(x2 + 2), how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?

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zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics @ganeshie8 @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I'm pondering...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we have to note that we have the subsequent inequality: \[\Large {x^2}  4 < 0,\quad {\text{if }}  2 < x < 2\]

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0honestly, im confused

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3it is a simple inequality

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0no i understand the inequality, im confused by how you got it

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! 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
 one year ago
Best ResponseYou've already chosen the best response.0otherwise any other would be out of bounds

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3no, it is a step in order to get rid the absolute value symbol

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now, we can apply the theorem af mean value

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so, 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
 one year ago
Best ResponseYou've already chosen the best response.0\[f'(c) = \frac{ f(b)  f(a) }{ b  a }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \frac{{f\left( 1 \right)  f\left( 0 \right)}}{{1  0}} = \frac{{3 \times 3  4 \times 2}}{1} = \frac{{6  8}}{1} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3sorry I have made a big error, you are right

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0no apologies necessary! thank you for your help

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3next 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
 one year ago
Best ResponseYou've already chosen the best response.3what do you get?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3then we have to solve this equation: \[ \Large  4x\left( {{x^2}  1} \right) = 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3it is not a simple equation

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0\[x^3 = \frac{ 1 }{ 2 }\] ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3how did you do to write that expression?

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0\[4x(x^21) = 1\]\[\frac{ 1 }{ 4x } = x^2\] \[\frac{ 1 }{ 4x } +1 = x^2\] \[1+1 = 4x^3\] \[ \frac{ 1 }{ 2 } = x^3\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I don't understand this step: \[\frac{1}{{  4x}} = {x^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3anyway it is not a solution of our equation

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, i forgot the 1

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0How would you get the proper solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I questioned wolfram alpha, here are the solutions:

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0So, 3 solutions in the interval

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we have two solutions, inside the interval [0,1]

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, my bad! I looked outside the interval

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3do you see them?

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Yes. Since one of them is in the negatives, it is not in the interval and does not matter to answer the question

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3yes! that's right!

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0I tried doing the same thing on this problem, but I keep getting no solutions: A particle moves along the xaxis with position function s(t) = xe^x. How many times in the interval [−5, 5] is the velocity equal to 0?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0velocity = (s(t))' take derivative and let it =0 to solve for x .
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