Name: Date:Write down the inverse function f^{-1}(x) , given:
| Question 1: f(x)=2 - 5cos(x) | Question 2: f(x)={\Large \frac {\sqrt{8x + 7}}{5}} |
| Question 3: f(x)={\Large \frac {9x^2 + 4}{5}} | Question 4: f(x)={\Large\frac{6}{x}} + 2 |
| Question 5: f(x)=\sqrt{\Large\frac{8x - 3}{7}} | Question 6: f(x)={\Large \frac {\sqrt{3x - 2}}{7}} |
| Question 7: f(x)={\Large\frac{x}{8}} + 3 | Question 8: f(x)=3 + 4sin(x) |
| Question 9: f(x)={\Large\frac{7sin(x)}{4}} | Question 10: f(x)={\Large\frac{4}{x}} - 3 |
| Question 11: f(x)={\Large\frac{x}{5}} - 6 | Question 12: f(x)=\sqrt{\Large\frac{6x + 7}{5}} |
| Question 13: f(x)={\Large\frac{6(x - 7)}{5}} - 5 | Question 14: f(x)={\Large\frac{6}{5}}(x - 9) + 3 |
| Question 15: f(x)={\Large \frac {9x^2 - 7}{2}} | Question 16: f(x)=cos(7x - 5) |
| Question 17: f(x)=tan(9x + 3) | Question 18: f(x)={\Large\frac{9tan(x)}{5}} |
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| Question 1: f^{-1}(x)=cos^{-1}{\Large(\frac{2-x}{5})} | Question 2: f^{-1}(x)={\Large\frac{25x^2 - 7}{8}} |
| Question 3: f^{-1}(x)={\Large \sqrt{\frac{5x - 4}{9}}} | Question 4: f^{-1}(x)={\Large\frac{6}{x - 2}} |
| Question 5: f^{-1}(x)={\Large\frac{7x^2 + 3}{8}} | Question 6: f^{-1}(x)={\Large\frac{49x^2 + 2}{3}} |
| Question 7: f^{-1}(x)=8(x - 3) | Question 8: f^{-1}(x)=sin^{-1}{\Large(\frac{x-3}{4})} |
| Question 9: f^{-1}(x)=sin^{-1}{\Large(\frac{4x}{7})} | Question 10: f^{-1}(x)={\Large\frac{4}{x + 3}} |
| Question 11: f^{-1}(x)=5(x + 6) | Question 12: f^{-1}(x)={\Large\frac{5x^2 - 7}{6}} |
| Question 13: f^{-1}(x)={\Large\frac{5(x + 5)}{6}} + 7 | Question 14: f^{-1}(x)={\Large\frac{5(x - 3)}{6}} + 9 |
| Question 15: f^{-1}(x)={\Large \sqrt{\frac{2x + 7}{9}}} | Question 16: f^{-1}(x)={\Large\frac{cos^{-1}(x) + 5}{7}} |
| Question 17: f^{-1}(x)={\Large\frac{tan^{-1}(x) - 3}{9}} | Question 18: f^{-1}(x)=tan^{-1}{\Large(\frac{5x}{9})} |