Consider the function f:R^ + [-9, ] given by f(x) = 5x2 + 6x - 9. Prove that f is invertible with f^ -1 (y)= 54+5y -3 5 .

f:R^ + [-9, ) given by f(x) = 5x^2 + 6x - 9 For any x, y ^ + f(x) = f(y) 5x^2 + 6x - 9 = 5y^2 + 6y - 9 5(x^2 - y^2) + 6(x - y) = 0 (x - y)[5(x + y) + 6] = 0 (x - y) = 0 [ 5(x+y)+6 0 as x, y ^ + ] x = y So, f is an injection. Let y be an arbitrary element of [-9, ). f(x) = y 5x^2 + 6x - 9 = y 25x^2 + 30x - 45 = 5y 25x^2 + 30x + 9 - 54 = 5y (5x + 3)^2 = 5y + 54 (5x+3)= 5y+54 x= 5y+54 -3 5 Now, y [-9, ) y -9 5y+54 9 5y+54 3 5y+54 -3 0 5y+54 -3 5 0 x 0 x ^ + Thus, for every y [-9, ) there exist x= 5y+54 -3 5 ^ + such that f(x) = y. So, f:R^ + [-9, ) is onto. Thus, f:R^ + [-9, ) is a bijection and hence invertible. Let f^ -1 denotes the inverse of f. Then, (fof^ -1 )(y) = y for all y [-9, ) f(f^ -1 (y)) = y for all y [-9, ) 5(f^ -1 (y))^2 + 6(f^ -1 (y)) - 9 = y for all y [-9, ) 25(f^ -1 (y))^2 + 30(f^ -1 (y)) - 45 = 5y for all y [-9, ) 25(f^ -1 (y))^2 + 30(f^ -1 (y)) + 9 = 5y + 54 for all y [-9, ) 5f^ -1 (y) + 3 ^2 = 5y + 54 for all y [-9, ) 5f^ -1 (y) + 3 = 5y+54 for all y [-9, ) f^ -1 (y)= 5y+54 -3 5

Consider the function f:R^ + [-9, ] given by f(x) = 5x2 + 6x - 9.…

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Question

Consider the function $\text{f}:\text{R}^{+}\rightarrow[-9,\infty]$ given by $f(x) = 5x2 + 6x - 9.$ Prove that $f$ is invertible with $\text{f}^{-1}\text{(y)}=\frac{\sqrt{54+5\text{y}}-3}{5}.$

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Answer

$\text{f}:\text{R}^{+}\rightarrow\ [-9,\infty)$ given by $f(x) = 5x^2 + 6x - 9$ For any $\text{x, y}\in\text{R}^{+}$
$f(x) = f(y)$
$\Rightarrow 5x^2 + 6x - 9 = 5y^2 + 6y - 9$
$\Rightarrow 5(x^2 - y^2) + 6(x - y) = 0$
$\Rightarrow (x - y)[5(x + y) + 6] = 0$
$\Rightarrow (x - y) = 0 [\because5(\text{x}+\text{y})+6\neq0\text{ as x, y}\in\text{R}^{+}]$
$\Rightarrow x = y$
So, $f$ is an injection.
Let $y$ be an arbitrary element of $[-9,\infty).$
$f(x) = y$
$\Rightarrow 5x^2 + 6x - 9 = y$
$\Rightarrow 25x^2 + 30x - 45 = 5y$
$\Rightarrow 25x^2 + 30x + 9 - 54 = 5y$
$\Rightarrow (5x + 3)^2 = 5y + 54$
$\Rightarrow(5\text{x}+3)=\sqrt{5\text{y}+54}$
$\Rightarrow\ \text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}$
Now, $\text{y}\in[-9,\infty)$
$\Rightarrow\ \text{y}\geq-9$
$\Rightarrow\ 5\text{y}+54\geq9$
$\Rightarrow\ \sqrt{5\text{y}+54}\geq3$
$\Rightarrow\ \sqrt{5\text{y}+54}-3\geq0$
$\Rightarrow\ \frac{\sqrt{5\text{y}+54}-3}{5}\geq0$
$\Rightarrow\ \text{x}\geq0\Rightarrow\ \text{x}\in\text{R}^{+}$
Thus, for every $\text{y}\in[-9,\infty)$ there exist $\text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}\in\text{R}^{+}$ such that $f(x) = y.$
$So, \text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is onto.
Thus, $\text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is a bijection and hence invertible.
Let $f^{-1}$ denotes the inverse of $f.$
Then,
$(fof^{-1})(y) = y$ for all $\text{y}\in[-9,\infty)$
$f(f^{-1}(y)) = y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 5(f^{-1}(y))^2 + 6(f^{-1}(y)) - 9 = y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) - 45 = 5y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) + 9 = 5y + 54$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow {5f^{-1}(y) + 3}^2 = 5y + 54$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 5f^{-1}(y) + 3 =\sqrt{5\text{y}+54}$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow\ \text{f}^{-1}(\text{y})=\frac{\sqrt{5\text{y}+54}-3}{5}$

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