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.

Answer 1

Solution:

As we know that a function is invertible(I.e. it's inverse is exist ) if and only if function is Bijective(one-one and onto)

Since it is defined from [0,∞)→ R

on putting few values from domain,we find that each has a different image ,thus function is one to one,but this is not onto because every element of range does not have pre-image in its Domain.

Or we can try to find its inverse,we will find that it is impossible to convert this function into x =f(y)

$f(x) = 9 {x}^{2} + 6x - 5 \\ \\ y = 9 {x}^{2} + 6x - 5 \\ \\ y + 5 = 9 {x}^{2} + 6x$
now further it can't be solved

Correction: If we convert the function into complete square,than only it can be convertible

$y = 9 {x}^{2} + 6x - 5 \\ \\ y = ( {3x)}^{2} + 2(3x)(1) + {1}^{2} - 5 - 1 \\ \\ y = ( {3x + 1)}^{2} - 6 \\ \\ y + 6 = ( {3x + 1)}^{2} \\ \\ \sqrt{y + 6} = (3x + 1) \\ \\ \sqrt{y + 6} - 1 = 3x \\ \\ x = \frac{ \sqrt{y + 6} - 1 }{3} \\ \\ x = f(y) = {f}^{ - 1}$

Now we can see that if function includes the value from -5 to infinity,than only function is onto,so it's range or co-domain must be [-5,∞)

Thus by this way we can convert the given function into convertible form.