Question

Consider f: R->R given by f(x) =5x +2
a) show that f is
invertible ?
b) show that f is one-one​

Answer 1

it has given that, f : R → R given by, f(x) = 5x + 2

To show : (a) f is invertible.

(b) f is one - one.

solution : any function, y = f(x) will be one one only if f(x₁) = f(x₂) ⇒x₁ = x₂ where x₁ , x₂ ∈ domain of function.

here f(x) = 5x + 2

so, f(x₁) = 5x₁ + 2

and f(x₂) = 5x₂ + 2

now, f(x₁) = f(x₂)

⇒5x₁ + 2 = 5x₂ + 2

⇒5x₁ = 5x₂

⇒x₁ = x₂

Therefore f is one one function.

now let's check function is onto or not.

condition of onto function : co - domain = Range.

here function, f(x) = 5x + 2

function is defined for all real value of x

for all x ∈ R, range of function, y ∈ R

here, co domain = Range

so, f is onto function.

we know, when any function is one one and onto, it will be invertible function.

Therefore f is invertible function.

Answer 2

Answer:

f is invertible

Step-by-step explanation:

please mark as brainliest answer