a.) h(x)= 8-|x+4

b. f(x)=x^2-6x+4

c. m(x)=14+21x-2x^2-3x^3
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1 Answer

a) h(x)=8-|x+4| has a maximum when x+4=0 that is, x=-4. So for x<-4 f(x) is increasing to that maximum, and for x>-4 it decreases from that maximum. Increase and decrease are linear.

b) f(x)=x2-6x+4=x2-6x+9-5=(x-3)2-5 has a minimum of -5 when x=3. For x<3 f(x) decreases to the minimum, and for x>3 it increases from the minimum.

c) m(x)=14+21x-2x2-3x3=7(2+3x)-x2(2+3x)=(7-x2)(2+3x)=(√7-x)(√7+x)(2+3x). This tells us that the zeroes are at x=±√7 and x=-⅔. To find where m(x) is increasing or decreasing we need to find the derivative of m(x).

m'(x)=21-4x-9x2=0, x=(-2±√193)/9 at the extrema.

When x=-1.7658 (approx) we have a minimum so, when x<-1.7658 m(x) is decreasing to the minimum.

There's a maximum when x=1.3214 (approx), so between -1.7658 and 1.3214 m(x) is increasing towards the maximum.

When x>1.3214 m(x) decreases continuously from the maximum.

by Top Rated User (1.1m points)

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