Question
Consider the function $\text{f}:\text{R}^{+}\rightarrow[-9,\infty]$ given by f(x) = 5x2 + 6x - 9. Prove that f is invertible with $\text{f}^{-1}\text{(y)}=\frac{\sqrt{54+5\text{y}}-3}{5}.$
For any
$\text{x, y}\in\text{R}^{+}$f(x) = f(y)
⇒ 5x2 + 6x - 9 = 5y2 + 6y - 9
⇒ 5(x2 - y2) + 6(x - y) = 0
⇒ (x - y)[5(x + y) + 6] = 0
⇒ (x - y) = 0 $[\because5(\text{x}+\text{y})+6\neq0\text{ as x, y}\in\text{R}^{+}]$
⇒ x = y
So, f is an injection.
Let y be an arbitrary element of $[-9,\infty).$
f(x) = y
⇒ 5x2 + 6x - 9 = y
⇒ 25x2 + 30x - 45 = 5y
⇒ 25x2 + 30x + 9 - 54 = 5y
⇒ (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 $\text{y}\in[-9,\infty)$
⇒ 5(f-1(y))2 + 6(f-1(y)) - 9 = y for all $\text{y}\in[-9,\infty)$
⇒ 25(f-1(y))2 + 30(f-1(y)) - 45 = 5y for all $\text{y}\in[-9,\infty)$
⇒ 25(f-1(y))2 + 30(f-1(y)) + 9 = 5y + 54 for all $\text{y}\in[-9,\infty)$
⇒ {5f-1(y) + 3}2 = 5y + 54 for all $\text{y}\in[-9,\infty)$
⇒ 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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