-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Inverse Functions Question (1 Viewer)
- Thread starter Shuuya
- Start date
DatAtarLyfe
Booty Connoisseur
- Joined
- Mar 10, 2015
- Messages
- 1,805
- Gender
- Female
- HSC
- 2016
Is it part i or ii you're stuck on?
Shuuya
Well-Known Member
- Joined
- Sep 3, 2015
- Messages
- 833
- Gender
- Undisclosed
- HSC
- 2016
Both --- I'm close to the answer in part (i), but I'm consistently getting it wrong
rand_althor
Active Member
- Joined
- May 16, 2015
- Messages
- 554
- Gender
- Male
- HSC
- 2015
i)Could I please have some help with these questions --- I keep doing something wrong :/
Thanks!
Last edited:
Shuuya
Well-Known Member
- Joined
- Sep 3, 2015
- Messages
- 833
- Gender
- Undisclosed
- HSC
- 2016
Thanks! Realised that I was making a really stupid mistake :/i)
 &= \frac{\left (\dfrac{\mathrm{d}}{\mathrm{d}x}3x \right )(9+x^2)- \left (\dfrac{\mathrm{d}}{\mathrm{d}x}9+x^2 \right )\times 3x}{(9+x^2)^2} + \frac{3}{9+x^2}\,\,(*)\\ &=\frac{3(9+x^2)-2x \times 3x}{(9+x^2)^2} + \frac{3(9+x^2)}{(9+x^2)^2} \\&=\frac{27+3x^2-6x^2}{(9+x^2)^2} + \frac{27+3x^2}{(9+x^2)^2} \\&=\frac{54}{(9+x^2)^2}\end{align*})
Shuuya
Well-Known Member
- Joined
- Sep 3, 2015
- Messages
- 833
- Gender
- Undisclosed
- HSC
- 2016
Got it, thanks! I was doing 1/17 instead of 1/27 that's why I couldn't get the answer -_-, an anti-derivative of $\frac{2}{\left(9+x^2\right)^2}$ is $F(x):=\frac{1}{27} \left(\frac{3x}{9+x^2}+ \tan^{-1} \frac{x}{3}\right)$. Hence the integral evaluates to $F\left(\sqrt{3} \right) -F(0) = F\left(\sqrt{3}\right)$, as $F(0)=0$. I'll leave you to evaluate $F\left(\sqrt{3}\right)$.$)