What is the inflection point on the graph f(x)=(x^2)/(1-x^2)?

Answer 1

f(x) = (x²)/(1-x²)
f'(x) = [(1-x²)(2x) - (x²)(2x)] / (1-x²)²
f'(x) = [(1-x² - x²)(2x)] / (1-x²)²
f'(x) = [(1-2x²)(2x)] / (1-x²)²
f'(x) = 2x / (1-x²)²
f²(x) = -2(3x² + 1) / (1-x²)^3
(3x² + 1) can't = 0, so f²(x) is never = 0, so there is NO inflection point.
Answer below is incorrect in its colculations.

Answer 2

f(x) = x^2 / (1 - x^2) - undefined at x = +/- 1, of course.

f'(x) = [2x (1-x^2) - x^2(-2x)] / (1-x^2)^2
= [2x - 2x^3 + 2x^3] / (1-x^2)^2
= 2x / (1-x^2)^2
= 0 <=> x = 0.

First Derivative Test is the easiest here: at 0 the derivative obviously goes from negative to zero to positive (since the denominator is 1 at this point), so this is a minimum. So there are no stationary inflection points.

f"(x) = (2(1-x^2)^2 - 2x.2.(1-x^2).(-2x)) / (1-x^2)^4
= (2 - 4x^2 + 2x^4 + 8x^2 - 8x^6) / (1-x^2)^4
= 0 when 8x^6 - 2x^4 - 4x^2 - 2 = 0
<=> (x^2 - 1)(8x^4 + 6x^2 + 2) = 0
The second term is always positive, and the first is zero only at x = +/- 1 where the function is not defined. So there are no inclined inflection points either. So it's a trick quesion; there are no inflection points.

Answer 3

first derive the funtion to locate f'(x) enable x^2 = h(x) and (x^2+a million) = g(x) using the product rule... f'(x) = h'(x)*g(x) + h(x)*g'(x) so f'(x) = 2x(x^2+a million) + x^2(2x) ... f'(x) = 2x((x^2+a million)+x^2) then prepare the product rule lower back letting h(x) = 2x and g(x) = ((x^2+a million)+x^2) f"(x) = 2((x^2+a million)+(x^2)) + 2x(2x+2x) so... f"(x) = (4x^2+2)+(8x^2) f"(x) = 12x^2 + 2 now inflection factors is whilst the 2d dervative equals 0... 0 = 2(6x^2 + a million) divide the two aspects by technique of two 0 = 6x^2 + a million -a million = 6x^2 -a million/6 = x^2 x = sq. root of -a million/6 so there are actually not any actual inflection factors (which means the function does not turn from concave as much as concave down or vice versa...the only inflection factors are plus or minus i circumstances the sqaure root of a million/6

Answer 4

You need to take the second derivative and apply it against the critical points to see what if any points of inflection there are.

f(x) = x²/(1 - x²)

Take the derivative to look for critical points.

f'(x) = [2x(1 - x²) - (-2x)x²]/(1 - x²)²
f'(x) = [2x - 2x³ + 2x³]/(1 - x²)²
f'(x) = 2x/(1 - x²)² = 0
x = 0

Take the second derivative to see if it is a point of inflection.

f'(x) = 2x/(1 - x²)²

f''(x) = [2(1 - x²)² - 2(-2x)(1 - x²)(2x)]/(1 - x²)^4
f''(x) = [2(1 - x²) + (4x)(2x)]/(1 - x²)³
f''(x) = [2 - 2x² + 8x²]/(1 - x²)³
f''(x) = 2(1 + 3x²)/(1 - x²)³

The numerator is always positive so f''(x) ≠ 0.

So there is no point of inflection.

Answer 5

1, -1, and then also whatever x equals in this equation: (x^2-2x-1)=0


take the second derivative, set it equal to zero

f(x)= x^2/(1-x^2)
f '(x)= 2x/(1-x^2)^2
f " (x) eventually = (1-x^2)(2(1-x^2)+4x), set both terms equal to zero, so 1-x^2=0 and 2(1-x^2)+4x) = 0, solving, x = 1, -1, and then the 2nd equation breaks down to (x^2-2x-1)=0 and i dont feel like solving it

Answer 6

Well this is what the graph looks like:

http://i31.photobucket.com/albums/c392/teekshi33/graphc.jpg

I think it would be -1, 0, and -1.

Answer 7

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