Question

Please help me with the below question.

Answer 1

Let y = f(x). By the fundamental theorem of calculus, differentiating f(x) twice yields

$f'(x) = \dfrac1{4 + x + 3x^2}$

$f''(x) = -\dfrac{1 + 6x}{(4 + x + 3x^2)^2}$

y is concave upward wherever f''(x) > 0. The denominator is positive for all real x, so the numerator determines the sign of f''(x). We have 6x + 1 = 0 when x = -1/6, and

• for x ∈ (-∞, -1/6), we have e.g. with x = -1, 6x + 1 = -5 < 0

• for x ∈ (-1/6, ∞), we have e.g. with x = 0, 6x + 1 = 1 > 0

Then y is concave upward on the interval (-1/6, ∞).