Prove that the function f:[0, ∞) → R given by f(x) = 9x
^{
2
}
+ 6x – 5 is not invertible. Modify the codomain of the function f to make it invertible, and hence find f
^{
–1
}
.

Question 13 (OR 1st question) Prove that the function f:[0, ∞) → R given by f(x) = 9x2 + 6x – 5 is not invertible. Modify the codomain of the function f to make it invertible, and hence find f–1 .
If f(x) is invertible
f(x) is one-one
f(x) is onto
First, let us check if f(x) is onto
Let f(x) = y
y = 9x2 + 6x – 5
0 = 9x2 + 6x – 5 – y
9x2 + 6x – 5 – y = 0
9x2 + 6x – (5 + y) = 0
Comparing equation with
ax2 + bx + c = 0
a = 9, b = 6 , c = – (5 + y)
x = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎
Putting values
x = (− 6 ± √(6^2 − 4(9) (−(5 + 𝑦)) ))/2(9)
x = (− 6 ± √(36 + 36(5 + 𝑦)))/18
x = (− 6 ± √(36(1 + (5 + 𝑦)) ))/18
x = (− 6 ± √(36(6 + 𝑦) ))/18
x = (− 6 ± 6 √((6 + 𝑦)))/18
x = 6[− 1 ± √((6 + 𝑦) )]/18
x = (− 1 ± √((6 + 𝑦) ))/3
So, x = (− 1 − √((6 + 𝑦) ))/3 or (− 1 + √((6 + 𝑦) ))/3
As x ∈ [0, ∞) , i.e., x is a positive real number
x cannot be equal to (−1 − √((6 + 𝑦) ))/3
Hence, x = (−1 + √((6 + 𝑦) ))/3
Since y ∈ R
For y = –6
x = (−1 + √((6 − 6) ))/3 = (−1)/3
Since x = (−1)/3 ∉ [0, ∞)
So, f is not invertible
To make it invertible,
x ≥ 0
(−1 + √((6 + 𝑦) ))/3 ≥ 0
–1 + √((6 + 𝑦) ) ≥ 0
√((6 + 𝑦) ) ≥ 0 + 1
√((6 + 𝑦) ) ≥ 1
6 + y ≥ 1
y ≥ 1 – 6
y ≥ –5
So, our range should be [–5, ∞)
Now, our function is
f: [0,∞) → [–5, ∞)
f(x) = 9x2 + 6x – 5
We need to first check if it is one-one and onto
Checking one-one
f (x1) = 9(x1)2 + 6x1 – 5
f (x2) = 9(x2)2 + 6x2 – 5
Putting f (x1) = f (x2)
9(x1)2 + 6x1 – 5 = 9(x2)2 + 6x2 – 5
9(x1)2 – 9(x2)2 + 6x1 – 6x2 = – 5 + 5
9(x1)2 – 9(x2)2 + 6x1 – 6x2 = 0
Now, our function is
f: [0,∞) → [–5, ∞)
f(x) = 9x2 + 6x – 5
We need to first check if it is one-one and onto
Checking one-one
f (x1) = 9(x1)2 + 6x1 – 5
f (x2) = 9(x2)2 + 6x2 – 5
Putting f (x1) = f (x2)
9(x1)2 + 6x1 – 5 = 9(x2)2 + 6x2 – 5
9(x1)2 – 9(x2)2 + 6x1 – 6x2 = – 5 + 5
9(x1)2 – 9(x2)2 + 6x1 – 6x2 = 0
(x1 – x2) = 0
⇒ x1 = x2
(3x1 + 3x2 + 2) = 0
⇒ 3x1 = –3x2 – 2
Since x ∈ [0, ∞)
i.e. x is always positive,
Hence 3x1 = – 3x2 – 2 is not true
Hence, if f (x1) = f (x2) , then x1 = x2
∴ f is one-one
Now,
Checking onto
f(x) = 9x2 + 6x – 5
Putting f(x) = y,
We get x = (−1 + √((6 + 𝑦) ))/3
Since for y ∈ [–5, ∞)
x ∈ [0, ∞)
Thus, f is onto
Since the function is one-one and onto
∴ It is invertible
Calculating inverse
For finding inverse, we put f(x) = y and find x in terms of y
We have done that while proving onto
x = (−1 + √((6 + 𝑦) ))/3
Let g(y) = (−1 + √((6 + 𝑦) ))/3
where g: [–5,∞ ) → [0, ∞)
So, inverse of f = f–1 = (−𝟏 + √((𝟔 + 𝒚) ))/𝟑

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.