Consider f : R+ → [−5, ∞), given by f(x) = 9x2 + 6x − 5.

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394.

Consider f : R+ → [−5, ∞), given by f(x) = 9x2 + 6x − 5. Show that f is invertible with f−1(y)open parentheses fraction numerator square root of straight y plus 6 end root minus 1 over denominator 3 end fraction close parentheses.

Hence Find
(i) f−1(10)
(ii) y if f−1(y)=43,

where R+ is the set of all non-negative real numbers.


f : R+ → [−5, ∞) given by f(x) = 9x2+ 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
Let f(x1)=f(x2)
⇒9x12+6x1−5=9x22+6x2−5
⇒9x12+6x1=9x22+6x2
⇒9x12−9x22+6x1−6x2=0
⇒9(x12−x22)+6(x1−x2)=0
⇒(x1−x2)[9(x1+x2)+6]=0
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
9x2+6x−5−y=0
or
9x2+6x−(5+y)=0 where x∈R+
rightwards double arrow space straight x space equals space fraction numerator negative 6 space plus-or-minus 4 left parenthesis 9 right parenthesis left parenthesis negative left parenthesis 5 plus straight y right parenthesis right parenthesis over denominator 2 left parenthesis 9 right parenthesis end fraction
straight x space equals space fraction numerator negative 6 plus-or-minus square root of 36 plus 4 left parenthesis 9 right parenthesis left parenthesis 5 plus straight y right parenthesis end root over denominator 18 end fraction space
space equals fraction numerator negative 6 plus-or-minus square root of 36 plus end root square root of left parenthesis 6 plus straight y right parenthesis end root over denominator 18 end fraction space
space equals fraction numerator negative 6 plus-or-minus square root of 6 squared plus end root square root of left parenthesis 6 plus straight y right parenthesis end root over denominator 18 end fraction space
space equals space fraction numerator negative 6 plus-or-minus 6 square root of left parenthesis 6 plus straight y right parenthesis end root over denominator 18 end fraction space
equals space fraction numerator negative 6 open square brackets negative 1 plus-or-minus square root of left parenthesis 6 plus straight y right parenthesis end root close square brackets over denominator 18 end fraction space
equals fraction numerator negative 1 space plus-or-minus space square root of left parenthesis 6 plus straight y right parenthesis end root over denominator 3 end fraction
So space comma straight x space space equals space fraction numerator negative 1 space plus space square root of left parenthesis 6 plus straight y right parenthesis end root over denominator 3 end fraction space or space fraction numerator negative 1 space minus space square root of left parenthesis 6 plus straight y right parenthesis end root over denominator 3 end fraction

As x∈R+ i.e., is a positive real number
x cannot be equal to
fraction numerator negative 1 minus square root of open parentheses 6 plus straight y close parentheses end root over denominator 3 end fraction space therefore comma
straight x space equals fraction numerator begin display style negative 1 plus square root of open parentheses 6 plus straight y close parentheses end root end style over denominator begin display style 3 end style end fraction
Since f: R+ →[-5,∞)
so y ∈ [-5,∞)
i.e y is greater than or equal to -5
i.e. y ≥-5
y+5  ≥0
⇒ Hence the value inside root is positive
Hence √y +6≥0
⇒ x≥0
Hence x is a real number which is greater than or equal to 0.



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