Question

Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer 1

$\bold{Answer :}$

Given that,

f (x) = 4x + 3

(i) Let us take two distinct elements x₁, x₂ in ℝ, the domain of f.

f(x₁) = 4x₁ + 3

f(x₂) = 4x₂ + 3

f(x₁) - f(x₂) = 4(x₁ - x₂) ≠ 0, since x₁ ≠ x₂

Since x₁ ≠ x₂ implies f(x₁) ≠ f(x₂), f is injective.

(ii) Let us take an arbitrary element y in the set ℝ, the co-domain of f; and let us examine if y has a pre-image x in the domain of f.

Then f(x) = y and therefore 4x + 3 = y

or, x = (y - 3)/4

Since y ∈ ℝ, (y - 3)/4 ∈ ℝ. Therefore y has a pre-image (y - 3)/4 in the domain of f. Since y is arbitrary, each element in the co-domain of f has a pre-image under f. Then f is surjective.

Since f is both injective and surjective, f is bijective and thus f is invertible.

Also, f⁻¹ = (x - 4)/3, by replacing y by x in (y - 3)/4

#$\bold{MarkAsBrainliest}$

Answer 2

It is given that f : R → R given by f (x) = 4x + 3
Let x,y ∈ R such that f(x) = f(y)
⇒ 4x +3 = 4y +3
⇒ 4x = 4y
⇒ x = y
therefore,  f is one- one function.

Now, for y ∈ R, Let y = 4x +3
$\implies x =\frac{y - 3}{4}$ ∈ R
⇒ for any y ∈ R, there exists x =$\frac{y-3}{4}$ ∈ R
such that, f(x) =$f\left(\begin{array}{c}\frac{y-3}{4}\end{array}\right)=4\frac{y-3}{4}+3=y$
⇒ F is onto function.
Since, f is one –one and onto
therefore , f is inversible.

now, f(x) = 4x + 3
f(x) = y = 4x + 3
y = 4x + 3
=> y - 3 = 4x
=> 4x = (y - 3)
=> x = (y - 3)/4
=> f(y) = (y - 3)/4
here f(y) is not other than inverse of f(x)
put x in place of y and write f(y) = f⁻¹(x)
e.g., f⁻¹(x) = (x -3)/4
hence, inverse of f(x) = (x-3)/4