\(y = f(x)\) likh kar \(x\) aur \(y\) interchange karenge, phir quadratic formula se \(y\) solve karenge. \(\pm\) mein se woh branch rakhenge jiska range original restricted domain se match kare. Inverse ka domain = original ki range.
(i) \(f(x) = x^2 - 3,\; x \le 0\) — \(x = y^2 - 3 \implies y^2 = x + 3 \implies y = \pm\sqrt{x+3}\). The inverse's range must be \((-\infty, 0]\), so keep \(y = -\sqrt{x+3}\). Range of \(f\): at \(x = 0\), \(f = -3\); as \(x \to -\infty\), \(f \to \infty\). So:
\(f^{-1}(x) = -\sqrt{x+3}\); Domain of \(f^{-1} = [-3, \infty)\); Range of \(f^{-1} = (-\infty, 0]\). (Check: \(f(-2) = 1\) and \(f^{-1}(1) = -2\) ✓.)
(ii) \(f(x) = x^2 + 6x + 4,\; x < -3\) — \(x = y^2 + 6y + 4 \implies y^2 + 6y + (4-x) = 0\). Quadratic formula:
\[
y = \frac{-6 \pm \sqrt{36 - 4(4-x)}}{2} = \frac{-6 \pm \sqrt{20+4x}}{2} = -3 \pm \sqrt{x+5}
\]
Range of \(f^{-1}\) must be \((-\infty, -3)\), so keep \(y = -3 - \sqrt{x+5}\). Vertex of \(f\) is \((-3, -5)\); for \(x < -3\), \(f\) decreases from \(\infty\) to \(-5\) (excluded), so Range of \(f = (-5, \infty)\).
\(f^{-1}(x) = -3 - \sqrt{x+5}\); Domain of \(f^{-1} = (-5, \infty)\); Range of \(f^{-1} = (-\infty, -3)\).
(iii) \(f(x) = 2x^2 - 8x + 11,\; x \ge 2\) — \(x = 2y^2 - 8y + 11 \implies 2y^2 - 8y + (11-x) = 0\):
\[
y = \frac{8 \pm \sqrt{64 - 8(11-x)}}{4} = \frac{8 \pm \sqrt{8x-24}}{4} = \frac{8 \pm 2\sqrt{2x-6}}{4} = \frac{4 \pm \sqrt{2x-6}}{2}
\]
Range of \(f^{-1}\) must be \([2, \infty)\), so keep the \(+\) branch. Vertex of \(f\): \(h = 2\), \(k = 8 - 16 + 11 = 3\); Range of \(f = [3, \infty)\).
\(f^{-1}(x) = \frac{4 + \sqrt{2x-6}}{2}\); Domain of \(f^{-1} = [3, \infty)\); Range of \(f^{-1} = [2, \infty)\).
(iv) \(f(x) = 3x^2 - 2x + 6,\; x \ge 5\) — \(x = 3y^2 - 2y + 6 \implies 3y^2 - 2y + (6-x) = 0\):
\[
y = \frac{2 \pm \sqrt{4 - 12(6-x)}}{6} = \frac{2 \pm \sqrt{12x-68}}{6} = \frac{1 \pm \sqrt{3x-17}}{3}
\]
Keep the \(+\) branch for range \([5, \infty)\). \(f(5) = 75 - 10 + 6 = 71\), and \(f\) increases for \(x \ge 5\), so Range of \(f = [71, \infty)\).
\(f^{-1}(x) = \frac{1 + \sqrt{3x-17}}{3}\); Domain of \(f^{-1} = [71, \infty)\); Range of \(f^{-1} = [5, \infty)\).
(v) \(f(x) = 2(x-3)^2 + 1,\; x \ge 3\) — \(x = 2(y-3)^2 + 1 \implies (y-3)^2 = \frac{x-1}{2} \implies y = 3 \pm \sqrt{\frac{x-1}{2}}\). Keep \(+\):
\(f^{-1}(x) = 3 + \sqrt{\frac{x-1}{2}}\); Domain of \(f^{-1} = [1, \infty)\); Range of \(f^{-1} = [3, \infty)\).
(vi) \(f(x) = -3(x+4)^2 - 5,\; x < -4\) — \(x = -3(y+4)^2 - 5 \implies (y+4)^2 = \frac{-(x+5)}{3} \implies y = -4 \pm \sqrt{\frac{-(x+5)}{3}}\). Range of \(f^{-1}\) must be \((-\infty, -4)\), so keep \(y = -4 - \sqrt{\frac{-(x+5)}{3}}\). Vertex \((-4, -5)\); for \(x < -4\), \(f\) rises from \(-\infty\) to \(-5\) (excluded): Range of \(f = (-\infty, -5)\).
\(f^{-1}(x) = -4 - \sqrt{\frac{-(x+5)}{3}}\); Domain of \(f^{-1} = (-\infty, -5)\); Range of \(f^{-1} = (-\infty, -4)\).
Answer: (i) \(f^{-1}(x) = -\sqrt{x+3}\), Dom \([-3,\infty)\), Ran \((-\infty,0]\) (ii) \(f^{-1}(x) = -3-\sqrt{x+5}\), Dom \((-5,\infty)\), Ran \((-\infty,-3)\) (iii) \(f^{-1}(x) = \frac{4+\sqrt{2x-6}}{2}\), Dom \([3,\infty)\), Ran \([2,\infty)\) (iv) \(f^{-1}(x) = \frac{1+\sqrt{3x-17}}{3}\), Dom \([71,\infty)\), Ran \([5,\infty)\) (v) \(f^{-1}(x) = 3+\sqrt{\frac{x-1}{2}}\), Dom \([1,\infty)\), Ran \([3,\infty)\) (vi) \(f^{-1}(x) = -4-\sqrt{\frac{-(x+5)}{3}}\), Dom \((-\infty,-5)\), Ran \((-\infty,-4)\)
