Points of inflection of (1-x^2)/x^3?

Question

What are the points of inflection of the function m(x) = (1-x^2)/x^3?

Answer 1

find where the 2nd derivative = 0 and the 3rd derivative does not equal 0

Answer 2

m'(x)=(x^2-3)/x^4

m"(x)=(-2x^2+12)/x^5

m"(x)=0 <=> -2x^2+12=0 <=> x=+-square root(6)

Answer 3

f(x) = (1 - x^2) / (x^3)

Take the second derivative.

f'(x) = [ (-2x)(x^3) - (1 - x^2)(3x^2) ] / [ x^6 ]

f'(x) = [ -2x^4 - 3x^2(1 - x^2) ] / [ x^6 ]

f'(x) = [ -2x^4 - 3x^2 + 3x^4 ] / [ x^6 ]

f'(x) = [ x^4 - 3x^2 ] / [ x^6 ]

f'(x) = [ x^2(x^2 - 3) ] / [ x^6 ]

f'(x) = [ x^2 - 3 ] / [ x^4 ]

f''(x) = [ (2x)(x^4) - (x^2 - 3)(4x^3) ] / [ x^8 ]

f''(x) = [ 2x^5 - 4x^3(x^2 - 3) ] / [ x^8 ]

f''(x) = [ 2x^5 - 4x^5 + 12x^3 ] / [ x^8 ]

f''(x) = [ 12x^3 - 2x^5 ] / [ x^8 ]

f''(x) = [ x^3(12 - 2x^2) ] / [ x^8 ]

f''(x) = [ 12 - 2x^2 ] / [ x^5 ]

Find the critical values for f''(x).  Make f''(x) = 0.

0 = [ 12 - 2x^2 ] / [ x^5 ]

Now, solve for x.  This is 0 when the numerator is 0.  

0 = 12 - 2x^2

2x^2 = 12

x^2 = 6

x = +/- sqrt(6)

Critical values are also for when f''(x) is undefined.  Therefore, 0 is also a critical value.

Test when f''(x) > 0 and when f''(x) < 0 by drawing a number line consisting of all critical values.

. . . . . . . . -sqrt(6) . . . . . . . .{0} . . . .  . . . . . sqrt(6) . . . . . . . . . . . .

Since f''(x) = [ 12 - 2x^2 ] / [ x^5 ], we want to test a single value in each region for positivity/negativity.

For the first region, test x = -100.  Then

f''(-100) = [ 12 - {something large} ] / [ something negative ]

= [ negative ] / [ negative ]

= [ positive ]

Mark the region as positive.

. . .{+}. . . . -sqrt(6) . . {-}. . . .{0} . . . . {+}. . . sqrt(6) . . . .{-}. . . . . .

For the second region,

Test x = -1.   Then

f''(1) = [ 12 - (-1)^2 ] / [ (-1)^5 ]

f''(1) = [ 12 - 1 ] / [ -1 ]

f''(1) = 11/(-1) = -11, which is negative.  Mark the region as negative.

For the third region, test x = 1.  Then f''(1) is positive.

For the 4th region, test x = 100.  We will get a negative number.

Conclusion:  In the negative regions, our function is concave down; in the positive regions, concave up.  At any alternation between critical values, we have an inflection point for the _defined_ critical values.

f(x) is concave up on (-infinity, -sqrt(6) ) U ( 0, sqrt(6) )

f(x) is concave down on ( -sqrt(6), 0 ) U ( sqrt(6), infinity )

We have inflection points at x = sqrt(6) and x = -sqrt(6).

But we require the y-coordinate as well.

f(sqrt(6)) = (1 - 6) / (6 sqrt(6) )

= (-5/6) (1/sqrt(6))

= (-5/6) (sqrt(6)/6)

= -5sqrt(6)/36

f(-sqrt(6)) = (1 - 6) / ( -6sqrt(6))

= 5sqrt(6)/36

So the inflection points are at ( sqrt(6) , -5sqrt(6)/36 ) and

( -sqrt(6) , 5sqrt(6)/36 )