Question

Consider f: R+ → [−5, ∞) given by f(x) = 9x^2 + 6x − 5. Show that f is invertible with f-1(y) = [ {(y + 6)^1/2 - 1} / 3 ].

Answer 1

any function f is $\bf{inversible.}$
i) f is one - one /injective mapping
ii) f is onto /surjective mapping

so, Let's check given function is inversible or not .

Given, f:R+→ [-5,∞) given by f(x) = 9x² + 6x - 5
Let x , y ∈ R+ such that f(x) = f(y)
e.g., 9x² + 6x - 5 = 9y² + 6y - 5
=> 9x² - 9y² + 6x - 6y = 0
=> 9(x - y)(x + y) + 6(x - y) = 0
=> (x - y)[9(x + y) + 6 ] = 0
because x ,y ∈ R+
so, [9(x + y) + 6 ] ≠ 0
therefore , x - y = 0 => x = y
hence, f is one - one .

any function is onto ,when co-domain= range.
co - domain ∈ [-5, ∞)
Let's find range of given function ,
f(x) = 9x² + 6x - 5
=> y = 9x² + 6x - 5
=> 9x² + 6x - 5 - y = 0
=> 9x² + 6x - (5 + y) = 0
=> x = [-6 ± √{(36 +36(5+y)}]/18
=> x = {-1 ± √(6 + y)}/3
because x ∈ R+ , x ≠ {-1 - √(6 + y)}/3
so, x = {-1 + √(6 + y)}/3
because x ∈ R+ e.g., x ≥ 0
=> {-1 + √(6 + y)}/3 ≥ 0
=> -1 + √(6 + y) ≥ 0
=> 6 + y ≥ 1
=> y ≥ -5 , hence range ∈ [-5, ∞ )
here you can see co-domain = range
therefore, f is onto function.

since f is one - one and onto.
therefore, f is inversible .

now, f(x) = 9x² + 6x - 5
=> y = 9x² + 6x - 5
=> 9x² + 6x - 5 - y = 0
=> x = [-6 ± √{36 + 36(5 + y)}]/18
=> x = [-1 ± √(6 + y)]/3
because x ∈ R+ , x ≠ {-1 - √(6 + y)}/3
so, x = {-1 + √(6 + y)}/3
hence, f⁻¹〈x〉 = {-1 + √(6 + x)}/3