Question

Let f: R → R, determine whether f is invertible, and if so, determine f

-1

, where f = {(x,

y)|2x+3y = 7}.​

Answer 1

SOLUTION

Let f: R → R , where f = {(x, y) | 2x+3y = 7}.

TO DETERMINE

whether f is invertible, and if so, determine

$\sf{ {f}^{ - 1} }$

EVALUATION

Here it is given that

f: R → R, , where f = {(x, y) | 2x+3y = 7}.

$\displaystyle\sf{2x + 3y = 7}$

$\displaystyle\sf{ \implies \: y = \frac{7 - 2x}{3} }$

$\displaystyle\sf{ \implies \: f(x) = \frac{7 - 2x}{3} }$

CHECKING FOR ONE TO ONE

Let a, b R such that

f(a) = f(b)

$\displaystyle\sf{ \implies \: \frac{7 - 2a}{3} = \frac{7 - 2b}{3} }$

$\displaystyle\sf{ \implies \: {7 - 2a} = {7 - 2b}}$

$\displaystyle\sf{ \implies \: a = b}$

So f is one to one

CHECKING FOR ONTO

Clearly every element in the codomain set has a preimage in the domain set

So f is onto

So f is bijective

So f is invertible

$\displaystyle\sf{Let \: \: {f}^{ - 1}(x) = y }$

$\displaystyle\sf{ \implies \: \: f(y) = x }$

$\displaystyle\sf{ \implies \: \frac{7 - 2y}{3} = x }$

$\displaystyle\sf{ \implies \: 7 - 2y = 3x }$

$\displaystyle\sf{ \implies \: - 2y = 3x - 7 }$

$\displaystyle\sf{ \implies \: 2y = 7 - 3x }$

$\displaystyle\sf{ \implies \: y = \frac{7 - 3x}{2} }$

$\displaystyle\sf{ \implies \: {f}^{ - 1}(x) = \frac{7 - 3x}{2} }$

FINAL ANSWER

$\displaystyle\sf{ {f}^{ - 1}(x) = \frac{7 - 3x}{2} }$

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