f(x)=(x+13)⋅ln(x+3) for x>-3 is concave upward on the interval (___,inf).

Question

f(x)=(x+13)⋅ln(x+3) for x>-3 is concave upward on the interval (___,inf).

Answer 1

A function is concave up when the second derivative > 0.

Take the first derivative by using the product rule:

f'(x)=(x+13)'⋅ln(x+3) + (x+13)⋅[ln(x+3)]'

f'(x)=1⋅ln(x+3) + (x+13)⋅1/(x+3)

f'(x)=ln(x+3) + (x+13)/(x+3)

Take the second derivative by using the quotient rule:

f"" = (f')' = [ln(x+3)]' + [(x+13)'(x+3) - (x+13)(x+3)']/(x+3)²

f"" = 1/(x+3) + [(x+3) - (x+13)]/(x+3)²

f"" = 1/(x+3) + -10/(x+3)²

f"" = (x+3)/(x+3)² + -10/(x+3)²

f'' = (x-7)/(x+3)²

To find when f(x) is concave up, set f'' > 0

(x-7)/(x+3)² > 0

But, since the denominator is always > 0 (it's a square), then (x-7)/(x+3)² > 0 when (x-7) > 0

so x>7

So the interval f(x) is concave up is (7,inf)