Find concavity 2nd derivative

Question

For what interval is f(x) = 1/x^2+1 concave down?

_____ < x < _____

Answer 1

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Answer 2

I'm going to assume that the function is:

Let's take the derivative. To do so, let's consider the function as (x² + 1)^-1 so we can use the power rule and the chain rule (as opposed to the quotient rule). So f '(x) = -2x(x² + 1)^-2. By the way, this is equal to zero at x = 0 so x = 0 is the only critical point for the function.

We must use the product rule to take the second derivative:

f '(x) = -2x(x² + 1)^-2 and, since the product rule says: (u•v)' = u'v + uv':

u = -2x

u' = -2

v = (x² + 1)^-2

v' = -2((x² + 1)^-3(2x) = -4x(x² + 1)^-3

So:

f ''(x) = (-2)(x² + 1)^-2 + (-2x)(-4x(x² + 1)^-3)

f ''(x) = -2(x² + 1)^-2 + 8x²(x² + 1)^-3

f ''(x) = -2(x² + 1)^-2[1 - 4x²(x² + 1)^-1]

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

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

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

f ''(x) = -2(1 - 3x²)/(x² + 1)³

f ''(x) = (-2 + 6x²)/(x² + 1)³

Setting this equal to zero:

(-2 + 6x²)/(x² + 1)³ = 0

The denominator cannot contribute to the left side of this equation being equal to zero so we can just look at the numerator:

-2 + 6x² = 0

6x² = 2

x² = 1/3

x = ±√(1/3) = ±√3/3

Set up a number line with these on it:

Plug in values in each of the 3 sections of this number line.

For the left side, let x = -1:

f ''(-1) = (-2 + 6(-1)²)/((-1)² + 1)³ = 4/8 which is positive, therefore the function is concave up on the interval containing -1, which is (-∞, -√3/3)

For the center section, let x = 0:

f ''(0) = (-2 + 6(0)²)/((0)² + 1)³ = -2/1 which is negative, therefore the function is concave down on the interval containing 0, which is (-√3/3, √3/3)

For the left side, let x = 1:

f ''(1) = (-2 + 6(1)²)/((1)² + 1)³ = 4/8 which is positive, therefore the function is concave up on the interval containing 1, which is (√3/3, ∞)

So the function is concave down only on (-√3/3, √3/3)