let ( f ) be the function defined by ( f(x)=2 x^{3}-3 x^{2}-12 x + 18 ). on which of the following intervals…

let ( f ) be the function defined by ( f(x)=2 x^{3}-3 x^{2}-12 x + 18 ). on which of the following intervals is the graph of ( f ) both decreasing and concave up?\na ( (-infty,-1) )\nb ( left(-1, \frac{1}{2}\right) )\nc ( (-1,2) )\nd ( left(\frac{1}{2}, 2\right) )\ne ( (2, infty) )

let ( f ) be the function defined by ( f(x)=2 x^{3}-3 x^{2}-12 x + 18 ). on which of the following intervals is the graph of ( f ) both decreasing and concave up?\na ( (-infty,-1) )\nb ( left(-1, \frac{1}{2}\right) )\nc ( (-1,2) )\nd ( left(\frac{1}{2}, 2\right) )\ne ( (2, infty) )

Answer

Explanation:

Step1: Find the first derivative

The first derivative (f'(x)) determines if the function is increasing or decreasing. Using the power rule ((x^n)' = nx^{n - 1}), for (f(x)=2x^{3}-3x^{2}-12x + 18), we have (f'(x)=6x^{2}-6x - 12=6(x^{2}-x - 2)=6(x + 1)(x - 2)). Set (f'(x)<0) (function is decreasing). (6(x + 1)(x - 2)<0). The critical points are (x=-1) and (x = 2). Using a sign - chart or test - values: the solution of (f'(x)<0) is (-1<x<2).

Step2: Find the second derivative

The second derivative (f''(x)) determines the concavity. Differentiate (f'(x)=6x^{2}-6x - 12) with respect to (x). Using the power rule, (f''(x)=12x-6 = 6(2x - 1)). Set (f''(x)>0) (function is concave up). (6(2x - 1)>0), which gives (2x-1>0) or (x>\frac{1}{2}).

Step3: Find the intersection of the intervals

We want the intersection of (-1<x<2) (from (f'(x)<0)) and (x>\frac{1}{2}) (from (f''(x)>0)). The intersection of (-1<x<2) and (x>\frac{1}{2}) is (\frac{1}{2}<x<2).

Answer:

D. ((\frac{1}{2},2))