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

- anonymous

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- Michele_Laino

the first derivative of your function is:
\[\Large f'\left( x \right) = \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}}\]

- Michele_Laino

am I right?

- anonymous

yes

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## More answers

- Michele_Laino

ok! Now we have to establish where that first derivative is positive

- Michele_Laino

So we have to solve this inequality:
\[\Large \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}} > 0\]

- Michele_Laino

what are the solutions of that inequality?

- anonymous

would you set the denominator to = 0? i got +-3

- 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

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

Oh okay that makes sense

- Michele_Laino

ok! Now what are the solutions of:
\[\Large - 2x > 0\]

- anonymous

it would be 0

- Michele_Laino

if I multiply, both sides of that inequality by -1, I get:
\[\Large 2x < 0\]
am I right?

- anonymous

yes

- Michele_Laino

ok! Now please, divide both sides of the last inequality by 2, what do you get?

- anonymous

x<0 (0/2)

- Michele_Laino

perfect!

- 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

increasing*

- Michele_Laino

so what can you conclude?

- anonymous

Okay, would my intervals be then increasing from (-infinity,0) and decreasingfrom (0,-infinity)?

- 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

your last statement means that x=0 is a point of local maximum for f(x), am I right?

- Michele_Laino

decreasing in (0, +infinity) and increasing in (-infinity, 0)

- 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

no, they are not. Those points don't belong to the domain of our function

- anonymous

i thought both infinities were (-)?

- anonymous

oh okay!

- Michele_Laino

at x=3 and x=-3 we have two vertical asymptotes

- 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

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

yes! that's right, we have an increasing function in:
(-infinity, 0)
and a decreasing function in:
(0, +infinity)

- anonymous

oh i just got it, i kept looking at the y values but its the x values that matter

- anonymous

so for my local max, it would just be x=0?

- Michele_Laino

yes! That's right!

- anonymous

and the function doesn't have a local min right?

- Michele_Laino

that's right, we have an unique point of local maximum for f(x)

- 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

here we can compute the second derivative of f(x)

- anonymous

6(x^2+3)/(x^2-9)^3

- 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

the same result! :)

- 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

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

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

now, since 1>0, then f ''(x)>0 when:
\[\Large {x^2} - 9 > 0\]
am I right?

- anonymous

yes

- anonymous

what about the power of 2 though?

- Michele_Laino

ok! what are the solutions of:
\[\Large {x^2} - 9 > 0\]?

- anonymous

you would get +- 3

- 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

oh okay!

- 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

yes!

- 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

please remember there is a theorem which states that a function is concave up in all points for which f ''(x) >0

- anonymous

okay! thank you so much!!

- Michele_Laino

:)

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