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