Consider f - R -5- - given by f x -9 x 2-6 x -5 show that f is ivnertible with f -1 y -y-6-1-3

Here, function f:R_+ → [ 5, ∞) is given as f(x)=9x^2 +6x 5. Let y be any arbitrary element of [ 5,∞). Let y=9x^2 +6x 5 ⇒ y=(3x+1)^2 1 5=(3x+1)^2 6 ⇒ (3x+1)^2=y+ ...

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Standard XII

Mathematics

Bijective Function

Consider f : ...

Question

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

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Solution

Here, function $f : R_{+} \to [- 5, \infty)$ is given as $f (x) = 9 x^{2} + 6 x - 5.$
Let y be any arbitrary element of $[- 5, \infty) .$
Let $y = 9 x^{2} + 6 x - 5$
$\Rightarrow y = (3 x + 1)^{2} - 1 - 5 = (3 x + 1)^{2} - 6 \Rightarrow (3 x + 1)^{2} = y + 6$
$\Rightarrow 3 x + 1 = \sqrt{y + 6} [a s y \geq - 5 \Rightarrow y + 6 \geq 0]$

$\Rightarrow x = \frac{\sqrt{y + 6} - 1}{3}$
Therefore, f is onto, thereby range $f = [- 5, \infty) .$
Let us define $g : [- 5, \infty) \to R_{+} a s g (y) = \frac{\sqrt{y + 6} - 1}{3}$
Now, $(g o f) (x) = g (f (x)) = g (9 x^{2} + 6 x - 5) = g ((3 x + 1)^{2} - 6)$
$= \frac{\sqrt{(3 x + 1)^{2} - 6 + 6 - 1}}{3} = \frac{3 x + 1 - 1}{3} = x$
and $(f o g) (y) = f (g (y)) = f (\frac{\sqrt{y + 6} - 1}{3}) = {[3 (\frac{\sqrt{y + 6} - 1}{3}) + 1]}^{2} - 6$
$= (\sqrt{y + 6})^{2} - 6 = y + 6 - 6 = y$
Therefore, $g o f = I_{R_{+}}$ and $f o g = I_{[- 5, \infty]}$
Hence, f is invertible and the inverse of f if given by
$f^{- 1} (y) = g (y) = \frac{\sqrt{y + 6} - 1}{3}$

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Consider f:R+→[−5,∞) given by f(x)=9x2+6x−5 show that f is ivnertible with f−1(y)=((√y+6)−13)

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