Question

Let A = (-3,-2,-1,1,2,3) and B = {1,4,9). If f:
AB defined by f (x)= x for all xe A
hen, range f = f (A) = {f (-3), f(-2), f (-1), $ (1) $ (2), f (3)}={1,4,9}=B.
.: f: AB is a surjection. Note that f is not an injection​

Answer 1

Answer:

A=R−{3}

B=R−{1}

f:A→B

f(x)=

x−3

x−2

f(x

1

)=f(x

2

)

x

1

−3

x

1

−2

=

x

2

−3

x

2

−2

(x

2

−3)(x

1

−2)=(x

2

−2)(x

1

−3)

x

1

x

2

−3x

1

−2x

2

+6=x

1

x

2

−3x

2

−2x

1

+6

−3x

1

−2x

2

=−3x

2

−2x

1

−x

1

=−x

2

x

1

=x

2

So, f(x) is one-one

f(x)=

x−3

x−2

y=

x−3

x−2

y(x−3)=x−2

yx−3y=x−2

yx−x=3y−2

x(y−1)=3y−2

x=

(y−1)

3y−2

f(x)=

x−3

x−2

=

y−1

3y−2

−3

y−1

3y−2

−2

=

y−1

3y−2−3(y−1)

y−1

3y−2−2(y−1)

=

3y−2−3y+3

3y−2−2y+2

=

−2+3

3y−2y

=y

f(x)=y

f(x) is onto.

So f(x) is bijective and invertible

f(x)=

x−3

x−2

y=

x−3

x−2

x=

y−3

y−2

x(y−3)=y−2

xy−3x=y−2

xy−y=3x−2

y(x−1)=3x−2

y=

x−1

3x−2

f

−1

(x)=

x−1

3x−2

Step-by-step explanation:

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