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