finding critical, inflection, extrema and concavity

Question

Let f(x) = (x^3)/(1 − x^2). 
Find 
(a) the first and second derivative of f 
(b) all critical points of f 
(c) all local extrema of f 
(d) all inflection points of f 
(e) where the graph of f is concave up and where it is concave down 

What I done so far: 
f'(x)=3x^2-5x^4
f"(x)=6-60x^3

 

 

Answer 1

a)

Apply first derivative both sides with respect to x.

Apply again(second) derivative both sides with respect to x.

Solution :

.

Answer 2

b)

To find the critical numbers , equate the first derivative equal to zero ⇒ f ' (x) = 0.

.

Answer 3

c)

Substitute critical points in f(x) to determine local extrema.

Solution :

Answer 4

d)

To find the inflection points , equate the second derivative equal to zero ⇒ f ''(x) = 0.

Solutions are x = 0, 1 and - 1

To find the y coordinate of inflection points substitute x = 0, 1, - 1 in original function.

At x = 0

The point is (0, 0)

Not defined.

Not defined.

Therefore, Inflection point is (0, 0).

Answer 5

e)

Inflection points are x = -1 , 0 , 1

The test intervals are (-∞ , -1) , (-1 , 0) , ( 0 , 1 ) and (1 , ∞).

 

In case of test points no need to calculate denominator.

As it consists square term denominator never affects sign of f '' (x).

 

Let test point x = - 2 in Interval (-∞, -1)

Let test point x = -0.5 in Interval (-1 , 0)

Let test point x = 0.5 in Interval (0 , 1)

Let test point x = 2 in Interval (1, ∞)

Interval    Test Value     Sign of f''(x)            Conclusion           

(-∞ , -1)     x = -2          Positive f''(x) > 0      Concave upward.

(-1 , 0)       x = - 0.5      Negative f''(x) < 0     Concave downward.

(0 , 1)        x = 0.5         Positive f''(x) > 0      Concave upward.

(1 , ∞)       x = 2           Negative f''(x) < 0    Concave downward.

Graph :