The largest open interval over which f is concave up for f(x) = 6x³ + 4x²- 3x + 2 is

Question

Quizplus · Accepted Answer

To find the intervals where the function is concave up, we need to find the second derivative and set it greater than 0. Taking the second derivative of the given function, we get:
f''(x) = 72xS1U1P9S1S1P0 + 24xS1U1P8S1S1P0 F1F1F1S1F1F1F10
This second derivative is a quadratic function of x with a positive leading coefficient. Therefore, it will be positive for all values of x except possibly at the roots. The roots can be found by setting the quadratic equal to zero:
72xS1U1P9S1S1P0 + 24xS1U1P8S1S1P0 F1F1F1S1F1F1F10 = 0
x(72xS1U1P8S1S1P0 + 24S1F1F1S1F1F1F10) = 0
x = 0 or x = -1/3
Note that we have used the fact that S1U1P9S1S1P0 = S1 + 11U1 + 11P9 + S1 + 11S1 + P0 = S1S1S1S1S1S1S1S1S1S1S1S1U1P9S1S1P0 and similarly for other subscripts.
Since the quadratic has a positive leading coefficient, the only possible change in concavity occurs at these roots. Therefore, the function is concave up for all values of x except possibly on the intervals (-∞,-1/3) and (0,∞). Since we want the largest open interval over which the function is concave up, we choose (-1/3,∞). Note that this interval is open at -1/3 because f'' is not defined at x = -1/3.

The Largest Open Interval Over Which F Is Concave Up