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 Pre-Calculus Pre-Calculus Math Forum

 November 26th, 2014, 09:21 PM #1 Newbie   Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0 Analytic Trig problem Hi, need some help from the math gods: find the inverse function f^-1 of each function f. Find the range of f and domain and range of f^-1. f(x) = -2cos(3x); 0≤x≤π/3 far as steps goes, I made it this far on my own y = -2cos(3x) x = -2cos(3y) -x/2 = -2cos(3y)/-2 (-x/2)^3 = (cos(3y))^3 then I dragged the 3 out of make 1/3 (or square root it) 1/3(-x/2) = cosy cosy = 1/3(-x/2) y = 1/3cos(-x/2) Is this right? If not, can you have a step by step process where I can understand easily? Thank you. Last edited by skipjack; November 27th, 2014 at 01:55 AM. November 26th, 2014, 09:51 PM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Your result is correct but your working is confused. $\displaystyle y=-2\cos(3x)$ $\displaystyle -\frac{y}{2}=\cos(3x)$ $\displaystyle \cos^{-1}\left(-\frac{y}{2}\right)=3x$ $\displaystyle x=\frac{\cos^{-1}\left(-\frac{y}{2}\right)}{3}$ $\displaystyle y^{-1}=\frac{\cos^{-1}\left(-\frac{x}{2}\right)}{3}$ November 27th, 2014, 02:02 AM   #3
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Quote:
 Originally Posted by p41s3r need some help
Do you need help with obtaining the domain and range for the inverse? November 27th, 2014, 12:57 PM   #4
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 Originally Posted by skipjack Do you need help with obtaining the domain and range for the inverse?
No, I know how to do those, but if its a tricky way to get them, I'll repost and I'll ask again, thank you and happy thanksgiving November 27th, 2014, 01:07 PM   #5
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 Originally Posted by greg1313 Your result is correct but your working is confused. $\displaystyle y=-2\cos(3x)$ $\displaystyle -\frac{y}{2}=\cos(3x)$ $\displaystyle \cos^{-1}\left(-\frac{y}{2}\right)=3x$ $\displaystyle x=\frac{\cos^{-1}\left(-\frac{y}{2}\right)}{3}$ $\displaystyle y^{-1}=\frac{\cos^{-1}\left(-\frac{x}{2}\right)}{3}$
Do you think, you can help me understand why I couldn't figure out the answer? If possible? Thanks for the reply, it really helps! Also, can you explain how your answer is y^-1=cos^-1(-x/2)/3 different from, y=1/3cos^-1(-x/2)? The answer I got, and I checked is from answers from the back, but I don't quite understand because I am not fluent in math! November 27th, 2014, 01:13 PM   #6
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 Originally Posted by skipjack Do you need help with obtaining the domain and range for the inverse?
You know what, I need clear instructions, because I'm not getting it, I don't have a strong background in math so I think I need help, so thank you so much! November 27th, 2014, 01:21 PM   #7
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Quote:
 Originally Posted by p41s3r . . . can you explain how your answer is y^-1=cos^-1(-x/2)/3 different from, y=1/3cos^-1(-x/2)?
You can ignore the "-1" and just use y. The "-1" was intended to show that the function is the inverse of the original function. So read it as:

$\displaystyle \text{The inverse of }-2\cos(3x)\text{ is }\frac{\cos^{-1}\left(-\frac{x}{2}\right)}{3}$

In short, they are the same thing.

Last edited by greg1313; November 27th, 2014 at 01:26 PM. Tags analytic, problem, trig Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post yogazen2013 Trigonometry 3 September 8th, 2013 04:17 AM tiba Geometry 3 June 10th, 2012 04:40 AM tiba Geometry 2 June 7th, 2012 03:03 PM davedave Complex Analysis 3 January 16th, 2010 04:33 PM frauleinedoctor Algebra 1 January 21st, 2009 05:43 AM

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