Expert: Abe Mantell - 1/5/2010

Question
I'm posed with the following problem:
let f(x) be the function defined by f(x) = k+12x+3x^2-2x^3 where k is a constant.
(a) On what interval is the function increasing?  Justify answer.
(b)if the relative maximum value of f is 4, what is the value of k?
(c) find the interval where the function is concave up.  Justify answer.
(d) find the relative minimum value of the function.

I'm not sure if I am supposed to use derivative tests to solve for the mins and maxs.  This is meant to be a non-calculator problem too.

Answer
Hello Yuki,

(a) If f'(x)>0, then f is increasing.  Thus, we need to solve for x when f'>0
.   f'=12+6x-6x^2>0 ==> 2+x-x^2>0 ==> x^2-x-2<0 ==> (x-2)(x+1)<0
==> x-2<0 AND x+1>0 ==> x<2 AND x>-1...thus, f is increasing for -1<x<2

(b) f'=0 ==> x=-1 or x=2 (see above for f'=0)...since f''=6-12x, we know the relative
.   maximum occurs at x=2 (since f''(2)<0).  Thus solve f(2)=4 for k.
.   ==> k+12*2+3*2^2-2*2^3=4 ==> k=-16

(c) concave up ==> f''>0 ==> 6-12x>0 ==> -12x>-6 ==> x<1/2

(d) from (b) we know the relative minimum must be at x=-1 (also since f''(-1)>0).
.   thus, it is f(-1) = -23

OK?

Abe