anonymous
  • anonymous
Find the intervals of increase and decrease of 1/x^2-9 b) find the coordinates of the local max and min Please help, I got x=0 and x=+-3 but im not sure what the intervals would be for the increase and decrease
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
the first derivative of your function is: \[\Large f'\left( x \right) = \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}}\]
Michele_Laino
  • Michele_Laino
am I right?
anonymous
  • anonymous
yes

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Michele_Laino
  • Michele_Laino
ok! Now we have to establish where that first derivative is positive
Michele_Laino
  • Michele_Laino
So we have to solve this inequality: \[\Large \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}} > 0\]
Michele_Laino
  • Michele_Laino
what are the solutions of that inequality?
anonymous
  • anonymous
would you set the denominator to = 0? i got +-3
Michele_Laino
  • Michele_Laino
no, since the denominator is always positive, then that fraction will be positive, if and only if also the numerator will be positive, namely: \[\Large - 2x > 0\]
Michele_Laino
  • Michele_Laino
the denominator: \[\Large {{{\left( {{x^2} - 9} \right)}^2}}\] is a square, which is always positive, and it is equal to zero at x=3 and x=-3 That means our first derivative doesn't exist at x=3 and x=-3. Also your function doesn't exist at x=3 and x=-3 for the same reason
anonymous
  • anonymous
Oh okay that makes sense
Michele_Laino
  • Michele_Laino
ok! Now what are the solutions of: \[\Large - 2x > 0\]
anonymous
  • anonymous
it would be 0
Michele_Laino
  • Michele_Laino
if I multiply, both sides of that inequality by -1, I get: \[\Large 2x < 0\] am I right?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
ok! Now please, divide both sides of the last inequality by 2, what do you get?
anonymous
  • anonymous
x<0 (0/2)
Michele_Laino
  • Michele_Laino
perfect!
Michele_Laino
  • Michele_Laino
that means our function is an incresing function when x<0 and a decreasing function when x>0: |dw:1435207357424:dw|
Michele_Laino
  • Michele_Laino
increasing*
Michele_Laino
  • Michele_Laino
so what can you conclude?
anonymous
  • anonymous
Okay, would my intervals be then increasing from (-infinity,0) and decreasingfrom (0,-infinity)?
Michele_Laino
  • Michele_Laino
that's right! Please keep in mind that at subsequent points: x=-3 and x=3, our function is not defined
Michele_Laino
  • Michele_Laino
your last statement means that x=0 is a point of local maximum for f(x), am I right?
Michele_Laino
  • Michele_Laino
decreasing in (0, +infinity) and increasing in (-infinity, 0)
anonymous
  • anonymous
thats what i have, but Im not sure if its right, would the -3 and 3 have anything to do with the local max and min?
Michele_Laino
  • Michele_Laino
no, they are not. Those points don't belong to the domain of our function
anonymous
  • anonymous
i thought both infinities were (-)?
anonymous
  • anonymous
oh okay!
Michele_Laino
  • Michele_Laino
at x=3 and x=-3 we have two vertical asymptotes
anonymous
  • anonymous
okay, i see it on the graph, you wrote decreasing (0,+infinity) and increasing (-infinity,0) why is it +infinity on the decreasing part? I thought it was - on both points
Michele_Laino
  • Michele_Laino
nevertheless those points (x=3, and x=-3) can not be considered as local maximum or minimu points since they don't check the definitions of local maximum or minimum points
Michele_Laino
  • Michele_Laino
yes! that's right, we have an increasing function in: (-infinity, 0) and a decreasing function in: (0, +infinity)
anonymous
  • anonymous
oh i just got it, i kept looking at the y values but its the x values that matter
anonymous
  • anonymous
so for my local max, it would just be x=0?
Michele_Laino
  • Michele_Laino
yes! That's right!
anonymous
  • anonymous
and the function doesn't have a local min right?
Michele_Laino
  • Michele_Laino
that's right, we have an unique point of local maximum for f(x)
anonymous
  • anonymous
okay! my next question is where its concave up and concave down. for concave up would it be (-3,-infinity)U(3,+infinity)?
Michele_Laino
  • Michele_Laino
here we can compute the second derivative of f(x)
anonymous
  • anonymous
6(x^2+3)/(x^2-9)^3
Michele_Laino
  • Michele_Laino
after a simple computation, I got: \[\Large f''\left( x \right) = 6\frac{{{x^2} + 3}}{{{{\left( {{x^2} - 9} \right)}^3}}}\]
Michele_Laino
  • Michele_Laino
the same result! :)
Michele_Laino
  • Michele_Laino
now the second derivative, can be rewritten as follows: \[\Large f''\left( x \right) = \frac{{6\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}\frac{1}{{\left( {{x^2} - 9} \right)}}\]
Michele_Laino
  • Michele_Laino
and we ahve to establish when f''(x) is positive, namely we have to study this inequality: \[\Large f''\left( x \right) > 0 \Rightarrow \frac{1}{{\left( {{x^2} - 9} \right)}} > 0\]
Michele_Laino
  • Michele_Laino
since: \[\Large \frac{{6\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}\] is always positive inside the domain of our function f(x)
Michele_Laino
  • Michele_Laino
now, since 1>0, then f ''(x)>0 when: \[\Large {x^2} - 9 > 0\] am I right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
what about the power of 2 though?
Michele_Laino
  • Michele_Laino
ok! what are the solutions of: \[\Large {x^2} - 9 > 0\]?
anonymous
  • anonymous
you would get +- 3
Michele_Laino
  • Michele_Laino
as you can see I factorized f ''(x) as follows: \[\Large f''\left( x \right) = \left[ {\frac{{6\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}} \right] \times \frac{1}{{\left( {{x^2} - 9} \right)}}\] now the quantity inside the square brackets is always positive, so we can neglect it when we have to solve the inequality f ''(x)>0
anonymous
  • anonymous
oh okay!
Michele_Laino
  • Michele_Laino
now the solution of this inequality: \[\Large {x^2} - 9 > 0\] is: \[\Large \left( { - \infty , - 3} \right) \cup \left( {3, + \infty } \right)\] right?
anonymous
  • anonymous
yes!
Michele_Laino
  • Michele_Laino
so we can conclude that our function is concave up inside that interval, namely in: \[\large \left( { - \infty , - 3} \right) \cup \left( {3, + \infty } \right)\] and concave down inside this interval: \[\Large \left( { - 3,3} \right)\]
Michele_Laino
  • Michele_Laino
please remember there is a theorem which states that a function is concave up in all points for which f ''(x) >0
anonymous
  • anonymous
okay! thank you so much!!
Michele_Laino
  • Michele_Laino
:)

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