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anonymous
 one year ago
Find the intervals of increase and decrease of 1/x^29
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
anonymous
 one year ago
Find the intervals of increase and decrease of 1/x^29 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

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3the first derivative of your function is: \[\Large f'\left( x \right) = \frac{{  2x}}{{{{\left( {{x^2}  9} \right)}^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! Now we have to establish where that first derivative is positive

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3So we have to solve this inequality: \[\Large \frac{{  2x}}{{{{\left( {{x^2}  9} \right)}^2}}} > 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3what are the solutions of that inequality?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would you set the denominator to = 0? i got +3

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3no, 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
 one year ago
Best ResponseYou've already chosen the best response.3the 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay that makes sense

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! Now what are the solutions of: \[\Large  2x > 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3if I multiply, both sides of that inequality by 1, I get: \[\Large 2x < 0\] am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! Now please, divide both sides of the last inequality by 2, what do you get?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3that means our function is an incresing function when x<0 and a decreasing function when x>0: dw:1435207357424:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so what can you conclude?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, would my intervals be then increasing from (infinity,0) and decreasingfrom (0,infinity)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3that's right! Please keep in mind that at subsequent points: x=3 and x=3, our function is not defined

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3your last statement means that x=0 is a point of local maximum for f(x), am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3decreasing in (0, +infinity) and increasing in (infinity, 0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats 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
 one year ago
Best ResponseYou've already chosen the best response.3no, they are not. Those points don't belong to the domain of our function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i thought both infinities were ()?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3at x=3 and x=3 we have two vertical asymptotes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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
 one year ago
Best ResponseYou've already chosen the best response.3nevertheless 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
 one year ago
Best ResponseYou've already chosen the best response.3yes! that's right, we have an increasing function in: (infinity, 0) and a decreasing function in: (0, +infinity)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i just got it, i kept looking at the y values but its the x values that matter

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for my local max, it would just be x=0?

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the function doesn't have a local min right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3that's right, we have an unique point of local maximum for f(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! my next question is where its concave up and concave down. for concave up would it be (3,infinity)U(3,+infinity)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3here we can compute the second derivative of f(x)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3after a simple computation, I got: \[\Large f''\left( x \right) = 6\frac{{{x^2} + 3}}{{{{\left( {{x^2}  9} \right)}^3}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3the same result! :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now 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
 one year ago
Best ResponseYou've already chosen the best response.3and 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
 one year ago
Best ResponseYou've already chosen the best response.3since: \[\Large \frac{{6\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2}  9} \right)}^2}}}\] is always positive inside the domain of our function f(x)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now, since 1>0, then f ''(x)>0 when: \[\Large {x^2}  9 > 0\] am I right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the power of 2 though?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! what are the solutions of: \[\Large {x^2}  9 > 0\]?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3as 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now the solution of this inequality: \[\Large {x^2}  9 > 0\] is: \[\Large \left( {  \infty ,  3} \right) \cup \left( {3, + \infty } \right)\] right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.3please remember there is a theorem which states that a function is concave up in all points for which f ''(x) >0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! thank you so much!!
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