Consider f- -5- - given by fx-9 x2-6 x-5- Show that f is invertible with f-1y-y-6-1-3-

F: R+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5. Let y be an arbitrary element of [−5, ∞). Let y = 9x2 + 6x − 5. ∴f is onto, thereby range f = [−5, ∞). Let us ...

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Standard XII

Mathematics

Differentiation of a Determinant

Consider f: �...

Question

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

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Solution

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

Let y be an arbitrary element of [−5, ∞).

Let y = 9x² + 6x − 5.

∴f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R₊ as

We now have:

∴and

Hence, f is invertible and the inverse of f is given by

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Similar questions

Q. Consider

f : R^{+} \in [- 5, \infty)

given by

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

. Show that

f

is invertible with

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

Consider $f : R_{+} \to [- 5, \infty)$ given by $f (x) = 9 x^{2} + 6 x - 5$ show that f is ivnertible with $f^{- 1} (y) = (\frac{(\sqrt{y + 6}) - 1}{3})$

Q. Consider f : R₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with

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

Q. Consider the function f : R⁺

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given by f(x) = 5x² + 6x

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9. Prove that f is invertible with f^{$-$ 1}(y) =

\frac{\sqrt{54 + 5 y} - 3}{5}

. [CBSE 2015]

Consider f: R₊→ [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f⁻¹ of given f by, where R₊ is the set of all non-negative real numbers.

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

f: R+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5. Let y be an arbitrary element of [−5, ∞). Let y = 9x2 + 6x − 5. ∴f is onto, thereby range f = [−5, ∞). Let us define g: [−5, ∞) → R+ as We now have: ∴and Hence, f is invertible and the inverse of f is given by

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

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

Let y be an arbitrary element of [−5, ∞).

Let y = 9x² + 6x − 5.

∴f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R₊ as

We now have:

∴and

Hence, f is invertible and the inverse of f is given by