differentiation

Question

Consider the function f(x) = -x³/3 - 1.5x² + 28x + 35.

 

What is the interval over which the function strictly concave?

Answer 1

f'(x) = -x² - 3x + 28
f''(x) = -2x - 3

inflection point when -2x - 3 = 0

-2x = 3

x = -1.5

 

f''(x) < 0 means concave downward

-2x - 3 < 0

-2x < 3

x > -1.5

 

so the curve is concave downward and permanently so on the interval (-1.5, ∞)