HELP please with this question!?

Question

Let f(x) = (x + 1)^2/(x – 1)^3. The f “(x) = 2x^2+20x+26/(x–1)^5. Over what intervals is the curve concave up?

a) x < -5 - 2√3 and -5 + 2√3 < x < 1
b) -5 -2√3 < x < -5 + 2√3 and x > 1
c) x > 1
d) -5 - 2√3 < x < -5 + 2√3

Answer 1

You get the critical points where the curve changes concavity by setting f''(x) to 0:

[2x² + 20x + 26] / (x-1)^5 = 0
x² + 10x + 13 = 0
x² + 10x + 25 = 12
(x + 5) = ±√12
x = -5 ± 2√3

the value of f''(x) will change signs at the critical points, and f(x) is concave up when f''(x) > 0, so picking x = -2 < -5 + 2√3,
f''(-2) = [8 - 40 + 26]/ -3^5 = -6/-243 > 0, so concave up for x between -5 - 2√3 and -5 + 2√3. Also, when x > 1 (x = 1, the vertical asymptote, is another critical point), f''(x) > 0, so concave up.

That makes your answer b).

Answer 2

The curve f(x) is concave up when the slope of the tangent line at any point is increasing, i.e., the second derivative f''(x) is positive. For f''(x) to be positive, its numerator and denominator must be both positive or both negative.

Both positive

2x^2 + 20x + 26 > 0 AND (x – 1)^5 > 0

2x^2 + 20x + 26 > 0

x < -5 - 2√3 or x > -5 + 2√3

AND

(x – 1)^5 > 0

x > 1

x < -5 - 2√3 or x > -5 + 2√3 AND x > 1

Leads to

x > 1


Both negative

2x^2 + 20x + 26 < 0 AND (x – 1)^5 < 0

2x^2 + 20x + 26 < 0

-5 - 2√3 < x < -5 + 2√3

AND

(x – 1)^5 < 0

x < 1

-5 - 2√3 < x < -5 + 2√3 AND x < 1

Leads to

-5 - 2√3 < x < -5 + 2√3

Answer is (b).

Answer 3

The answer is b) as the sign depends on
2x^2+20x+26 =2(x-(-5-2sqrt3))*(x(-(-5+sqrt3)) and on(x-1)
the sign of f´´ is

----(-5-2sqrt3)+++++(-5+2sqrt3
-------(1)+++++++++