1. The function f(x)=(x^24x5)/(x+1) is defined everywhere except at x = 1. If po

ID: 2872577 • Letter: 1

Question

1. The function f(x)=(x^24x5)/(x+1) is defined everywhere except at x = 1. If possible, define f (x) at 1 so that it becomes continuous at 1.

a) f(1)=5

b) f(1)=0

c) Not possible because there is an infinite discontinuity at the given point.

d) Not possible because there is a jump discontinuity at the given point.

e) f(1)=6

2. Give the values of A and B for the function f(x) to be continuous at both x = 1 and x = 2.

f(x)=AxB -->x1, 1x --> 1<x<2, Bx^2A --> x2

a) A = 1 and B = 1

b) A = 3 and B = 1

c) A = 2 and B = 0

d) A = 2 and B = 2

e) A = 2 and B = 1

1. Lets find the limit of f(x) at x=-1

lim(x-->-1) (x^24x5)/(x+1) =lim(x-->-1) (x+1)(x-5)/(x+1) =lim(x-->-1) (x-5) =-1-5= -6

As the limit exists we can make it continous by defining f(-1)=-6

2. If f(x) is continous at x=1 then

A*1-B=-1*1

A-B= -1 equation (1)

If f(x) is continous at x=2 then

-1*2=B*2^2-A

-2=4B-A equation (2)

From equation 1 and 2 we get

3B=-3

B=-1

A-B=-1

A-(-1)=-1

A+1=-1

A=-2 and B=-1

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.