Let f - N - R be a function defined as fx - 4x2 - 12x - 15- Show that f - N - S- where S is the range of f- is invertible- Find the inverse of f-

Calculate g : S → N Let f (x) = y y = 4x^2 nbsp; + 12x + 15 4x^2 nbsp; +12x + 15 y = 0 x = b ±√(b^2 4ac)2a Putting values, x = 12 ±√(12^2 4(4)(15 y) ...

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Standard XII

Chemistry

Heat Capacity

Let f : N → R...

Question

Let $f : N \to R$ be a function defined as $f (x) = 4 x^{2} + 12 x + 15$ . Show that $f : N \to S$ , where $S$ is the range of $f$ , is invertible. Find the inverse of $f$ .

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Solution

Calculate $g : S \to N$
Let $f (x) = y$
$y = 4 x^{2} + 12 x + 15$
$4 x^{2} + 12 x + 15 - y = 0$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Putting values,
$x = \frac{- 12 \pm \sqrt{12^{2} - 4 (4) (15 - y)}}{2 (4)}$
$x = \frac{- 12 \pm \sqrt{144 - 16 (15 - y}}{8}$
$= \frac{- 12 \pm \sqrt{16 (9 - 15 - y)}}{8}$
$= \frac{- 12 \pm \sqrt{16 (y - 6)}}{8}$
$= \frac{- 12 \pm \sqrt{16} \sqrt{y - 6}}{8}$
$= \frac{- 12 \pm \sqrt{4^{2}} \sqrt{y - 6}}{8}$
$= \frac{- 12 \pm 4 \sqrt{y - 6}}{8}$
$= \frac{4 [- 3 \pm \sqrt{y - 6}]}{8}$
$= \frac{- 3 \pm \sqrt{y - 6}}{2}$
So, $x = \frac{- 3 + \sqrt{y - 6}}{2}$ or $\frac{- 3 - \sqrt{y - 6}}{2}$
As $x \in N, x$ is a positive real number
$x$ can't be equal to $\frac{- 3 - \sqrt{y - 6}}{2}$
Hence, $x = \frac{- 3 + \sqrt{y - 6}}{2}$
Let $g (y) = \frac{- 3 + \sqrt{y - 6}}{2},$ where $g : S \to N$

Solve to prove $g o f = x = I_{N}$
$g o f (x) = g (f (x))$
$\Rightarrow g o f (x) = g (4 x^{2} + 12 x + 15)$
$\Rightarrow g o f (x) = \frac{- 3 + \sqrt{4 x^{2} + 12 x + 15 - 16}}{2}$
$\Rightarrow g o f (x) = \frac{- 3 + \sqrt{4 x^{2} + 12 x + 9}}{2}$
$\Rightarrow g o f (x) = \frac{- 3 \sqrt{(2 x)^{2} + 3^{2} + 2 (2 x) \times 3}}{2}$
$\Rightarrow g o f (x) = \frac{- 3 + \sqrt{(2 x + 3)^{2}}}{2}$
$\Rightarrow g o f (x) = \frac{- 3 + 2 x + 3}{2}$
$\Rightarrow g o f (x) = \frac{2 x}{2} = x$
Hence, $g o f = x = I_{N}$ .

Solve for prove $f o g = y = I_{y}$ .
$f o g (x) = f (g (x))$
$\Rightarrow f o g (x) = f (\frac{- 3 + \sqrt{y - 6}}{2})$
$\Rightarrow f o g (x) = 4 {(\frac{- 3 + \sqrt{y - 6}}{2})}^{2} + 12 (\frac{- 3 + \sqrt{y - 6}}{2}) + 15$
$\Rightarrow f o g (x) = 4 \frac{(- 3 + \sqrt{y - 6})^{2}}{4} + 6 (- 3 + \sqrt{y - 6} + 15$
$\Rightarrow f o g (x) = (- 3 + \sqrt{y - 6})^{2} - 18 + 6 \sqrt{y - 6} + 15$
$\Rightarrow f o g (x) = (- 3)^{2} + (\sqrt{y - 6})^{2} - 6 \sqrt{y - 6} - 18 + 6 \sqrt{(6 + y)} + 15$
$\Rightarrow f o g (x) = 9 + y - 6 - 18 + 15$
$\Rightarrow f o g (x) = y$
Hence, $f o g = y = I_{s}$
So, $g o f = I_{N} & f o g = I_{s}$ .
$\Rightarrow f$ is invertible. and inverse of $f (g (y) = \frac{- 3 + \sqrt{y - 6}}{2}$

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Let f:N→R be a function defined as f(x)=4x2+12x+15. Show that f:N→S, where S is the range of f, is invertible. Find the inverse of f.

Let $f : N \to R$ be a function defined as $f (x) = 4 x^{2} + 12 x + 15$ . Show that $f : N \to S$ , where $S$ is the range of $f$ , is invertible. Find the inverse of $f$ .