Name: Date:Write down the inverse function f^{-1}(x) , given:
| Question 1: f(x)={\Large\frac{9}{x}} + 4 | Question 2: f(x)={\Large \frac {7x^2 + 3}{2}} |
| Question 3: f(x)=\sqrt{\Large\frac{6x - 5}{7}} | Question 4: f(x)={\Large \frac {7x^2 - 3}{2}} |
| Question 5: f(x)={\Large\frac{4tan(x)}{7}} | Question 6: f(x)={\Large\frac{4(x + 6)}{3}} - 8 |
| Question 7: f(x)={\Large \frac {\sqrt{2x + 3}}{7}} | Question 8: f(x)={\Large\frac{x}{2}} - 8 |
| Question 9: f(x)={\Large\frac{6cos(x)}{5}} | Question 10: f(x)={\Large \frac {\sqrt{7x - 5}}{3}} |
| Question 11: f(x)=\sqrt{\Large\frac{3x + 7}{5}} | Question 12: f(x)={\Large\frac{x}{9}} + 3 |
| Question 13: f(x)=5 + 3sin(x) | Question 14: f(x)=3 - 7tan(x) |
| Question 15: f(x)={\Large\frac{5}{3}}(x + 7) + 2 | Question 16: f(x)={\Large\frac{3}{x}} - 2 |
| Question 17: f(x)=cos(7x + 2) | Question 18: f(x)=tan(6x - 8) |
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| Question 1: f^{-1}(x)={\Large\frac{9}{x - 4}} | Question 2: f^{-1}(x)={\Large \sqrt{\frac{2x - 3}{7}}} |
| Question 3: f^{-1}(x)={\Large\frac{7x^2 + 5}{6}} | Question 4: f^{-1}(x)={\Large \sqrt{\frac{2x + 3}{7}}} |
| Question 5: f^{-1}(x)=tan^{-1}{\Large(\frac{7x}{4})} | Question 6: f^{-1}(x)={\Large\frac{3(x + 8)}{4}} - 6 |
| Question 7: f^{-1}(x)={\Large\frac{49x^2 - 3}{2}} | Question 8: f^{-1}(x)=2(x + 8) |
| Question 9: f^{-1}(x)=cos^{-1}{\Large(\frac{5x}{6})} | Question 10: f^{-1}(x)={\Large\frac{9x^2 + 5}{7}} |
| Question 11: f^{-1}(x)={\Large\frac{5x^2 - 7}{3}} | Question 12: f^{-1}(x)=9(x - 3) |
| Question 13: f^{-1}(x)=sin^{-1}{\Large(\frac{x-5}{3})} | Question 14: f^{-1}(x)=tan^{-1}{\Large(\frac{3-x}{7})} |
| Question 15: f^{-1}(x)={\Large\frac{3(x - 2)}{5}} - 7 | Question 16: f^{-1}(x)={\Large\frac{3}{x + 2}} |
| Question 17: f^{-1}(x)={\Large\frac{cos^{-1}(x) - 2}{7}} | Question 18: f^{-1}(x)={\Large\frac{tan^{-1}(x) + 8}{6}} |