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Let $f : \mathrm{N\rightarrow N}$ be a function defined as $f(x) = 4x^2 +12x + 15$ . Show that $f : \mathrm{N\rightarrow S}$ is invertable ( Where S is the range of $f$ ). Find the inverse of $f$ and hence find $f ^{-1}(31)$ and $f ^{-1}(87)$ .

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Here $f : \mathrm{N\rightarrow N}$ , $f(x) = 4x^2 +12x + 15$

Let y be an arbitrary element of range of function $f$ . Then $y = 4x^2 +12x + 15$ , for some $x$ in N, which implies that $y = (2x+3)^2 + 6$

This gives $x = \frac{\sqrt{y-6}- 3}{2}$ as $y \geq 6$

Let us define $g: S\rightarrow N$ by $g(y) = \frac{\sqrt{y-6}- 3}{2}$

Now, $g\mathrm{o}f = g(f(x)) = g(4x^2+ 12x + 15)$

$= g((2x+3)^2 + 6) = \frac{\sqrt{((2x+3)^2 +6) -6} - 3}{2} = x$

And $f \mathrm{o} g(y) =f(g(y)) = f ( \frac{\sqrt{y-6}- 3}{2} ) = \left (2 \left (\frac{\sqrt{y-6}- 3}{2} \right ) + 3 \right ) ^2 + 6 = y$

Hence $g \mathrm{o} f = I_N$ and $f \mathrm{o} g = I_S$

This implies that $f$ is invertible with $f^{-1} = g$

S, $f^{-1} = \frac{\sqrt{y-6}- 3}{2}$ , i.e $f^{-1}(x) = \frac{\sqrt{x-6}- 3}{2}$

Now, $f^{-1}(31) = \frac{\sqrt{31-6}- 3}{2} = 1$ , $f^{-1}(87) = \frac{\sqrt{87-6}- 3}{2} = 3$

Posted by

Ravindra Pindel

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