Analysis of Functions Calculus

Question

Find the intervals of increase and decrease and the coordinates of the points of extrema:

a) f(x)=x+1/x^2

b) F(x)= (x²-1)(x+1)²

^

Answer 1

a)

Vertical asymptote at x=0

Solutions: x+1 = 0

x = -1

Sign table:

interval sample test value x f(x)

===============================================

x<-1 -2 -1/4

-1<x<0 -1/2 2

x>0 1 2

f'(x) = [x^2 - (2x)(x+1)]/ (x^4)

= [ x^2 - 2x^2 - 2x]/ (x^4)

= [ -x^2 - 2x]/x^4

= -x(x+2)/x^4

= -(x+2)/x^3

slopes are vertical at x=0

Critical points: x=-2 ---> (-2,-1/4)

Sign table:

interval sample test value x f(x)

=================================================

x<-2 -3 1/9

-2<x<0 -1 1

x>0 1 -3

decreases (-infinity,-2)

increases (-2,0)

decreases (0,infninity)

absolute max (-2, -1/4)

(B)

f(x) = (x^2-1)(x+1)^2

= (x+1)(x-1)(x+1)^2

= (x+1)^3(x-1)

Solutions:

x=-1 and x=1

y-intercept (0,-1)

sign chart

interval sample test value x f(x)

x<-1 -2 3

-1<x<1 0 -1

x>1 2 27

f'(x) = (x+1)^3 + 3(x+1)^2(x-1)

= (x+1)^2 [ x+1 + 3(x-1)]

= (x+1)^2 (4x-2)

= 2(x+1)^2 (2x-1)

horizontal tangents at x=-1 and x=1/2

sign chart

interval sample test value x f(x)

x<-1 -2 -10

-1 < x < 1/2 0 -2

x>1/2 1 48

decreases from (-infinity,1/2)

increases from (1/2,infinity)

f(1/2) = (1/4 - 1)(3/2)^2

= (-3/4)(9/4)

= -27/16

absolute min at (1/2,-27/16)