Question

Verify whether the function
f:R—>R defined by f(x) = 4x+3 is one-one ,onto and Bijective.
Justify your answer.​

Answer 1

f:R→R

f(x)=4x+3

if({x}^{1} )=f({x}^{2} )

4({x}^{1} ) +3=4({x}^{2} ) +3

({x}^{1} )= ({x}^{2} )

f(x) is one-one

f(x)=4x+3

y=4x+3

y−3=4x

$x= \frac{y - 3}{4}$

put x in

f(x)

$f(x)=4(\frac{y - 3}{4} ) + 3$

y−3+3=y

f(x) is onto

So ,

f(x) is invertible since the function is bijective.

f(x)=4x+3

x=4y+3

4y=x−3

$y = \frac{x - 3}{4}$

${f}^{ - 1} (x) = \frac{x - 3}{4}$

Answer 2

Step-by-step explanation:

f:R→R

f(x)=4x+3

if({x}^{1} )=f({x}^{2} )

4({x}^{1} ) +3=4({x}^{2} ) +3

({x}^{1} )= ({x}^{2} )

f(x) is one-one

f(x)=4x+3

y=4x+3

y−3=4x

x= \frac{y - 3}{4}x=

4

y−3

put x in f(x)

f(x)=4(\frac{y - 3}{4} ) + 3f(x)=4(

4

y−3

)+3

y−3+3=y

f(x) is onto

So ,

f(x) is invertible since the function is bijective.

f(x)=4x+3

x=4y+3

4y=x−3

y = \frac{x - 3}{4}y=

4

x−3

{f}^{ - 1} (x) = \frac{x - 3}{4}f

−1

(x)=

4

x−3