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.

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Solution

$f : [0, \infty) \to R$

domain of function= $D_{f} = [0, \infty)$ and co-domain of function is $C . D = R$

Now,

$f (x) = 9 x^{2} + 6 x - 5 = (9 x^{2} + 6 x + 1) - 6$

$⟹ f (x) = (3 x + 1)^{2} - 6$

Since $x \geq 0 ⟹ 3 x + 1 \geq 1$

$(3 x + 1)^{2} \geq 1 ⟹ (3 x + 1)^{2} - 6 \geq - 5$

$⟹ f (x) \geq - 5$

Range of function $R_{f} = [- 5, \infty) \neq C . D$

Since the function is not onto, it is not invertible as invertible function must be both one-one and onto.

Moreover this is a quadratic equation and not even is one-one.

$y = (3 x + 1)^{2} - 6$

$⟹ 3 x + 1 = \sqrt{y + 6}$

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

Hence $f^{- 1} (x) = \frac{- 1 + \sqrt{x + 6}}{3}$

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defined by

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

, where R is negative real numbers. Show that

f

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f : [0, \infty) \to R

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

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f : [0, \propto) \to R

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

is not invertible.

Q. 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

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Let f:[0,∞)→R be a function defined by f(x)=9x2+6x−5. Prove that f is not invertible and then find its inverse.

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.