how to find the intervals of concavity and points of inflection

Question

f(x) = x² / x²-4

i did the quotient rule and got

f '(x) = 8x² / (x² - 4)²

f '(x) = -8x (x²-4)^-2

and then i got

f" (x) = 16x (x²-4)^-3 (2x)

= 32x²(x²-4)^-3

or f " (x) = 32x² / (x² - 4)³

i need help finding the intervals and points of inflection, is my work good so far?

Answer 1

f(x) = x²/(x² - 4)

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

u' = 2x and v' = 2x

So f '(x) = [2x(x² - 4) - x²(2x)]/(x² - 4)² = (2x³ - 8x - 2x³)/(x² - 4)² = -8x/(x² - 4)²

f '(x) = -8x/(x² - 4)²

f ''(x) = (u'v - uv')/v² where u = -8x and v = (x² - 4)²

That means u' = -8 and v' (using the chain rule) = 2(x² - 4)(2x) = 4x(x² - 4) = 4x³ - 16x

So f ''(x) = [(-8)(x² - 4)² - (-8x)(4x³ - 16x)]/(x² - 4)⁴

f ''(x) = [-8(x⁴ - 8x² + 16) + 8x(4x³ - 16x)]/(x² - 4)⁴

f ''(x) = [-8x⁴ + 64x² - 128 + 32x⁴ - 128x²]/(x² - 4)⁴

f ''(x) = (24x⁴ - 64x² - 128)/(x² - 4)⁴

f ''(x) = 8(3x⁴ - 8x² - 16)/(x² - 4)⁴

f ''(x) = 8(x² - 4)(3x² + 4)/(x² - 4)⁴

f ''(x) = 8(3x² + 4)/(x² - 4)³

Setting f ''(x) equal to zero to find the points of inflection we get:

8(3x² + 4)/(x² - 4)³ = 0

The denominator cannot contribute to the rational expression being equal to zero therefore, we can ignore the denominator:

8(3x² + 4) = 0

3x² + 4 = 0

3x² = -4

x² = -4/3

The solutions are imaginary numbers so there are no possible points of inflection.

However, the concavity CAN change at points where f ''(x) DNE and that would be where the denominator of f ''(x) is zero. So, look at (x² - 4)³ being equal to zero:

(x² - 4)³ = 0

[(x + 2)(x - 2)]³ = 0

x = 2 and x = -2

So we can break up f(x) into intervals with those boundaries:

So let's try a point to the left of -2, how about -3?

The concavity at x = -3 is defined by f ''(-3) = 8(3(-3)² + 4)/((-3)² - 4)³ = 1.984 a positive number therefore f(x) is concave up on (-∞, -2).

Let's try a point between -2 and 2, how about zero?

The concavity at x = 0 is defined by f ''(0) = 8(3(0)² + 4)/((0)² - 4)³ = -1/2 a negative number therefore f(x) is concave down on (-2, 2).

So let's try a point to the right of 2, how about 3?

The concavity at x = 3 is defined by f ''(3) = 8(3(3)² + 4)/((3)² - 4)³ = 1.984 a positive number therefore f(x) is concave up on (2, ∞).

Answer 2

f(x) = x²/(x² - 4) = 1 + 4/(x² - 4) = 1 + 4(x²- 4)^-1;

f'(x) = - 8x(x²- 4)^-2 = - 8x/(x² - 4)²;

f''(x) = -8((x² - 4)² - 2(x² - 4)·2x·x)/(x² - 4)⁴ = -8(x² - 4 - 4x²)/(x² - 4)³ = 8(3x² + 4)/(x² - 4)³.

There is know inflection points because equation f''(x) = 0 has no real solutions.

Now f''(x) > 0 if x² - 4 > 0, so x < -2 or x > 2 and (- ∞, - 2), (2, ∞) intervals where graph concave up

f''(x) < 0 if x² - 4 < 0; - 2 < x < 2. So, (-2, 2) interval where graph concave down