find the point of inflection on the graph and describe the concavity

Question

Find the point of inflection of the graph of the function. (If an answer does not exist, enter DNE.)

f(x) = x

3 − x

(x, y) =

 

 

Describe the concavity. (Enter your answers using interval notation.)

concave upward
concave downward

Answer 1

For f(x) = x(3 - x)^1/2 we must use the product rule (for f(x) = u•v), f '(x) = u'v + uv')

u = x

u' = 1

v = (3 - x)^1/2

v' = -1/2(3 - x)^-1/2

So f '(x) = (3 - x)^1/2 - (x/2)(3 - x)^-1/2

The FIRST PART of the second derivative then is:

(-1/2)(3 - x)^-1/2 using the power rule and chain rule.

The SECOND PART of the second derivative is again found using the product rule where:

u = x/2

u' = 1/2

v = (3 - x)^-1/2

v' = 1/2(3 - x)^-3/2

or 1/2(3 - x)^-1/2 + (x/2)(1/2)(3 - x)^-3/2 so, putting those together we get:

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

f ''(x) = -(3 - x)^-1/2 - (x/4)(3 - x)^-3/2

f ''(x) = -4(3-x)/(4(3-x)^3/2) - x/(4(3-x)^3/2)

f ''(x) = (-12 + 3x)/(4(3-x)^3/2)

Setting this equal to zero, since the denominator cannot contribute to it equaling zero, we can just say:

-12 + 3x = 0 or x = 4

So we would normally think our point of inflection is x = 4 HOWEVER, in looking at the denominator, we can see that to get a real number, x must be less than 3. In fact, in the original function we see that the domain is x < 3

So when x < 3, we see that f '' will always be negative.

Therefore, the point of inflection DNE and the function is always concave downward over its domain (i.e., concave down on (-∞, 3) )