Consider f: `R_+ -> [-5, oo]` given by `f(x) = 9x^2 + 6x - 5`. Show that f is invertible with `f^(-1) (y) ((sqrt(y + 6)-1)/3)`
Hence Find
1) `f^(-1)(10)`
2) y if `f^(-1) (y) = 4/3`
where R+ is the set of all non-negative real numbers.
Solution
f: `R_+ -> [-5, oo]` given by `f(x) = 9x^2 + 6x - 5`
To show: f is one-one and onto.
Let us assume that f is not one-one.
Therefore there exist two or more numbers for which images are same.
For x1, x2 ∈ R+ and x1 ≠ x2
Since x1 and x2 are positive,
9(x1 + x2) + 6 > 0
∴ x1 − x2 = 0⇒ x1 = x2
Therefore, it contradicts our assumption.
Hence the function f is one-one.
Now, let is prove that f is onto.
A function f : X → Y is onto if for every y ∈ Y, there exist a pre-image in X.
f(x) = 9x2 + 6x −5
= 9x2 + 6x + 1 - 6
=(3x + 1)2 - 6
Now, for all x ∈ R+ or [0,∞), f(x) ∈ [−5, ∞)
∴ Range = co-domain.
Hence, f is onto.
Therefore, function f is invertible.
Now, let y = 9x2 + 6x − 5
1) `f^(-1) (10) = (sqrt(10+6)-1)/3 = (4-1)/3 = 1`
2) `f^(-1) (y) = 4/3`
`:.(sqrt(y + 6) -1)/3 = 4/3`
`=> sqrt(y + 6)-1 = 4`
`=> sqrt(y+6) = 5`
`=> y + 6 = 25`
=> y = 19
