Use interval notation to indicate where f(x) f ( x ) is concave up and down.

Question

if f(x)=2/(x^2-4)

Use interval notation to indicate where f(x) f ( x ) is concave up and down.

Answer 1

f(x) = 2/(x² – 4)

Use the Second Derivative Test to determine concave up or down.

                

Using the formula: if f(x) = u(x)*v(x), then f’(x) = u(x)*v’(x) + v(x)*u’(x), with u(x) = 2, and v(x) = (x² – 4)^-1 and the chain rule for 1/(x² – 4)

f’(x) = 2*(2x)(-1)/(x² – 4)² + 0 = -4x/( x² – 4)²

Using the same formula as above, with u(x) = -4x and v(x) = (x² – 4)^-2

f’’(x) = -4x*(-2)(2x)(x² – 4)^-3  - 4*(x² – 4)^-2  

                                                      

        = 16x²/(x² – 4)³ - 4/(x² – 4)²

        = (16x² – 4 x² + 16)/ (x² – 4)³

        = (12x² + 16)/ (x² – 4)³

From the above, it is clear that:

1)     The function and its derivatives are undefined if x = ±2, so any interval on either side of ±2 must be open at ±2 (i.e. does not include x=±2).

2)     f(x) is concave upward wherever it is positive => wherever f’’(x) = (12x² + 16)/ (x² – 4)³ > 0  

3)     f(x) is concave downward wherever it is positive => wherever f’’(x) = (12x² + 16)/ (x² – 4)³ < 0  

4)     The denominator is < 0 whenever x < 2

Since the denominator is to an odd power, f’’(x) < 0 whenever |x| < 2. Therefore, we have 3 intervals to look at:

Interval 1: x < -2. Then, x² – 4 > 0 => f’’(x) > 0 and therefore f(x) is concave upwards in the interval [-∞, -2).

Interval 2: |x| < 2. Then, x² – 4 < 0 => f’’(x) < 0 and therefore f(x) is concave downwards in the interval (-2,+2)

Interval 3: x > 2. Then, x² – 4 > 0 => f’’(x) > 0 and therefore f(x) is concave upwards in the interval (2,+∞]