Question

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?
a ( (-infty,-1) )
b ( left(-1, \frac{1}{2}
ight) )
c ( (-1,2) )
d ( left(\frac{1}{2}, 2
ight) )
e ( (2, infty) )

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

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

Answer:

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