i need help curve sketching this function

Question

y = 1+x² / 1-x²

this is what i have so far:

a. domain

D: (-∞, - 1) u (-1, 1) u (1, ∞), {x|x = 1, -1}

b. intercepts

there are no x-intercepts as -1 = x²

y = 1 therefore (0, 1)

c. symmetry

the function is even as f(-x) = f(x)

d. asymptotes

V.A are x = 1 and x = -1

H.A is y = -1

the rest i need help with!

e. intervals of increase or decrease

first derivative

y ' = 4x / (1 + x)² (1 - x)²

i do not know how to get the critical numbers, thus this prevents me from continuing the rest of the steps. what are the steps so i know?

f. maximum and minimum points

i cannot find the max. and min. points as i do not know how to get the critical numbers from the previous step

g. intervals of concavity

unable to find

h. points of inflection

unable to find

Answer 1

Good work so far. The first derivative requires you to use the quotient rule which goes like this:

For f(x) = u/v, f '(x) = (u'v - uv')/v²

In this case u = 1 + x² so u' = 2x. v = 1 - x² so v' = -2x

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

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

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

f '(x) = 4x/(1 - x²)²

The critical points are where f '(x) = 0 or where it DNE. So those would be x = 0, x = 1, x = -1

To find where the function is increasing or decreasing, divide a number line up into pieces divided by the critical points and then plug in a value (any convenient value) somewhere on each interval and plug it into the f ' function. If you get a positive, the function is increasing on that interval, if you get a negative, the function is decreasing on that interval. For instance, for the interval (-∞, -1) I'll selected x = -2. Plugging it in to f '(x) = 4x/(1 - x²)² I get f '(-2) = -8/9 which is negative, so the function is decreasing on (-∞, -1). Plugging in x = -0.5 for the interval (-1, 0) gives f '(-0.5) = -32/9 which is negative, so the function is decreasing on (-1, 0). Now do the same on (0, 1) and (1, ∞).

To find the maximum and minimum points, use the work above. If you find a function is decreasing on one side of a critical point and increasing on the other side of a critical point, and the function exists at the critical point, then that critical point is a local minimum.

To find the intervals of concavity, you need the second derivative.

f ''(x) = (u'v - uv')/v² where u = 4x and v = (1 - x²)²

So u' = 4 and v' = 2(1 - x²)(-2x) = (-4x)(1 - x²) = 4x³ - 4x

Put the derivative together like this:

f ''(x) = [4(1 - x²)² - (4x)(4x³ - 4x)]/(1 - x²)⁴

f ''(x) = [4(1 - 2x² + x⁴) - 16x⁴ + 16x²]/(1 - x²)⁴

f ''(x) = [4 - 8x² + 4x⁴ - 16x⁴ + 16x²]/(1 - x²)⁴

f ''(x) = [4 + 8x² - 12x⁴]/(1 - x²)⁴

f ''(x) = (-4)[3x⁴ - 2x² - 1]/(1 - x²)⁴

f ''(x) = (-4)(3x² +1 )(x² - 1)/(1 - x²)⁴

f ''(x) = (4)(3x² +1 )(1 - x²)/(1 - x²)⁴

f ''(x) = (4)(3x² +1 )/(1 - x²)³

Find the points where f '' = 0 or DNE. There are no points where it = 0 so there are no points of inflection. But there are points where the concavity changes. f '' DNE when x = 1 or x = -1. Divide the number line up on those intervals and find out which one has f '' > 0 and that will be where the function is concave up. Same with f '' < 0 where it will be concave down.