Trisarahtops
  • Trisarahtops
Medal!! Use the graph of f(x) = |x(x^2 − 1)| to find how many numbers in the interval [0.5, 0.75] satisfy the conclusion of the Mean Value Theorem. 1 2 4 None
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Trisarahtops
  • Trisarahtops
@Michele_Laino
nikeboi101
  • nikeboi101
its answer 1
Trisarahtops
  • Trisarahtops
how do u know?

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nikeboi101
  • nikeboi101
Did you mean |x(x^2-1)| ? f(x) is continuous for all real x f(x) is differentiable for all x ≠ −1, 0, or 1 f(x) is continuous on interval [1/2, 3/4] f(x) is differentiable on interval (1/2, 3/4) On the interval [1/2, 3/4] f(x) = −x(x^2−1) = −x^3+x f'(x) = −3x^2 + 1 (f(3/4)−f(1/2))/(3/4−1/2) = (21/64−3/8)/(1/4) = −3/16 Now we find all values of x on interval (1/2, 3/4) so that f'(x) = −3/16 −3x^2 + 1 = −3/16 3x^2 = 19/16 x^2 = 19/48 x = √(19/48) = 0.62915287 ---> ok [We ignore negative root, since it is not in the given interval] so yea its answer 1 @trisarahtops
nikeboi101
  • nikeboi101
@Trisarahtops
Trisarahtops
  • Trisarahtops
is that right @Michele_Laino
nikeboi101
  • nikeboi101
sure double checking is good
Michele_Laino
  • Michele_Laino
If I explicit the absolute value, I can write this: \[f\left( x \right) = \begin{array}{*{20}{c}} {x\left( {{x^2} - 1} \right),\quad x \in \left( { - 1,0} \right) \cup \left( {1, + \infty } \right)} \\ {x\left( {1 - {x^2}} \right),\quad x \in \left( { - \infty , - 1} \right) \cup \left( {0,1} \right)} \end{array}\]
Trisarahtops
  • Trisarahtops
so 2?
Michele_Laino
  • Michele_Laino
inside the interval \((1/2,3/4)\) we have two definitions for \(f(x)\)
Michele_Laino
  • Michele_Laino
so we have to do the computations separately, each computation for each subinterval
Michele_Laino
  • Michele_Laino
here is the first computation: \[\frac{{f\left( 1 \right) - f\left( {1/2} \right)}}{{1 - \left( {1/2} \right)}} = 1 - 3{c^2}\] where I have used: \[f\left( x \right) = x\left( {1 - {x^2}} \right)\]
Trisarahtops
  • Trisarahtops
f(1)-f(3/4) / 1-(3/4)
Michele_Laino
  • Michele_Laino
after a simple computation I get this: \[c = \sqrt {\frac{{19}}{{48}}} \simeq 0.63 < 1\] as we can see such value of \(c\) is greater than 0.5, and it is less than 1, so it is an acceptable value
Michele_Laino
  • Michele_Laino
next we have to do this computation: \[\frac{{f\left( {3/4} \right) - f\left( 1 \right)}}{{\left( {3/4} \right) - 1}} = 1 - 3{c^2}\] where: \[f\left( x \right) = x\left( {1 - {x^2}} \right)\]
nikeboi101
  • nikeboi101
so 1 is the answer
Michele_Laino
  • Michele_Laino
please wait, I'm doing the remaining computation after a simple computation, I get:
nikeboi101
  • nikeboi101
ok
Michele_Laino
  • Michele_Laino
I got this: \[c = \sqrt {\frac{{149}}{{128 \cdot 3}}} \simeq 0.623 < 3/4\] being such value less than \(3/4\), it can not be considered an acceptable value, so I confirm the answer of @nikeboi101
Trisarahtops
  • Trisarahtops
Thank you for clearing that up :D
Michele_Laino
  • Michele_Laino
:)
nikeboi101
  • nikeboi101
c:
Trisarahtops
  • Trisarahtops
you too ;)

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