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Consider F: `R_+ -> [-5, Infinite]` Given by `F(X) = 9x^2 + 6x - 5`. Show that F is Invertible with `F^(-1) (Y) ((Sqrt(Y + 6)-1)/3) Find Fpower(-1)(10) Where R+ Is the Set of All Non-negative Real Numbers. - Mathematics

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.

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Solution

f: `R_+ -> [-5, oo]` given by `f(x) = 9x^2 + 6x - 5`

To show: 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)- 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

  Is there an error in this question or solution?
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