Prove that the function $f [0, \infty) \to R$ given by $f (x) = 9 x^{2} + 6 x - 5$ is not the codomain of the function f to make it invertible, and hence find $f^{'}$

Open in App

Solution

$f (x) = 9 x^{2} + 6 x - 5$
The min value of $f (x)$ is at $x = - b / a = \frac{- 6}{9} = - \frac{2}{3}$
$f (- 2 / 3) = 9 \times \frac{4}{9} + 6 \times \frac{2}{3} - 5 4 + 4 - 5 = 3$
The function has no value less than $3$ but $R$ contains values less than $3$
So $f$ is not invertible.
$f^{'} (x) = \frac{d}{d x} (9 x^{2} + 6 x - 5) = 18 x + 6$

Suggest Corrections

Similar questions

Q. Prove that the function

f : [0, \infty) \to R

given by

f (x) = 9 x^{2} + 6 x - 5

is not invertible. Modify the codomain of the function f to make it invertible, and hence find

f^{- 1} .

Q. Prove that the function

f : [0, \propto) \to R

given by

f (x) = 9 x^{2} + 6 x - 5

is not invertible.

Let f:[0, infinity) to R be a function defined by f(x)= 9x² +6x-5. Proove that f is not invertible.Modify ,only the codomain of f to make f invertible and then find it's inverse .Also , find f^-1(43)

Q. Let

f : [0, \infty) \to R

be a function defined by

f (x) = 9 x^{2} + 6 x - 5

. Prove that

f

is not invertible and then find its inverse.

Q. Let

f : [0, \infty) \to R

be a function defined by

f (x) = 9 x^{2} + 6 x - 5

. Prove that f is not invertible. Modify, only the codomain of f to make f invertible and then find its inverse.

OR

Let

*

be a binary operation defined on

Q \times Q

(a, b) * (c, d) = (a c, b + a d)

, where Q is the set of rational numbers. Determine whether

*

is commutative and associative. Find the identity element for

*

and the invertible elements of

Q \times Q

Join BYJU'S Learning Program

Prove that the function f[0,∞)→R given by f(x)=9x2+6x−5 is not the codomain of the function f to make it invertible, and hence find f′

f(x)=9x2+6x−5The min value of f(x) is at x=−b/a=−69=−23f(−2/3)=9×49+6×23−54+4−5=3The function has no value less than 3 but R contains values less than 3So f is not invertible.f′(x)=ddx(9x2+6x−5)=18x+6

Prove that the function $f [0, \infty) \to R$ given by $f (x) = 9 x^{2} + 6 x - 5$ is not the codomain of the function f to make it invertible, and hence find $f^{'}$