# Calculus

Compute (f^-1) (2) if f(x) = 7x + 3cos(x) + 2sin(x).... We tried solving it in the form (f^-1)(a) = (1/f'(f^-1(a))... or is this a u-sub problem? We are trying to figure out whose way is right with one getting the answer as 1/2 and the other 1/10... we are getting mixed up with the f'(x) and the signs for cos and sin.... or we might both be completely wrong! Please show steps if possible, if not, at least the correct answer. Thanks!

1. 👍 0
2. 👎 0
3. 👁 115
1. well, it's easy enough to see whether an answer is right.

If f^-1(2) = x, then
f(x) = 2

clearly not.

It appears you want (f^-1)'(2)
f'(x) = 7-3sinx+2cosx
f'(2) = -3sin2+2cos2

(f^-1)'(2) = 1/f'(2)

1. 👍 0
2. 👎 0
posted by Steve
2. f^-1(x) definition is usually the inverse function
if y = 7x + 3cos(x) + 2sin(x)
substitute x for y and y for x and solve for y

x = 7 y + 3 cos y + 2 sin y

if x = 2

we want x = 2

2 = 7 y + 3 cos y + 2 sin y
y x
0, 3
pi/6 , 7.26
-pi/6 , -.067
- pi/12 , +.54

so somewhere round -pi/6

1. 👍 0
2. 👎 0
posted by Damon
3. I apologize! I forgot the prime mark after (f^-1)'(2). Thank you for catching that, you clearly know better what I ment than I did! So for clarification to compute (f^-1)'(2) if f(x) = 7x + 3cos(x) + 2sin(x), it would be 1/(-3sin(2) + 2cos(2)? That's the answer? Because the way I am doing it is...
f'(x) = 7 - 3cos(x) - 2cos(x) = 2 f(?) = 2
= 7 - 3cos(0) - 2cos(0) = 2
= 7-5 =2 ---------> 2 = 2
so (f^-1)(2) = 0
1 / 7- 3cos(0) - 2cos(0) = 1 / 7-3-2 = 1/2
So i am getting 1/2 as the answer. my friend has similar work but signs are switched around and he is getting 1/10. Or are we both just doing this completely wrong and neither of our answers are right? Thanks for your help!

1. 👍 0
2. 👎 0
posted by Robin
4. where did the 0 come from?
also you cannot have cos and cos in f'.

1. 👍 0
2. 👎 0
posted by Steve

## Similar Questions

1. ### trig

simplify: (2sin(2pie/7))/(3cos(3pie/14)

asked by Aya on December 7, 2011
2. ### Math

particles move at x=2sin(t), y=3cos(t),z=t at any t>0. What is the accelaration and jerk

asked by Anonymous on October 1, 2015
3. ### trig

What kind of reflections are the following trig functions? y = 3cos(x-1) y = sin(-3x+3) y = -2sin(x)-4

asked by Joshua on April 21, 2008
4. ### math

Find the area of the region enclosed between y=2sin(x) and y=3cos(x) from x=0 to x=0.7π. Hint: Notice that this region consists of two parts.

asked by Josh on April 27, 2019
5. ### Trig

what is the amplitude, period, phase shift, and vertical shift of each equation? 1. y= A cos(B(x-3))+4 2. y=-2sin(2piX/3) 3. y=4cos(3pi(x+3)) 4. y= -3cos(2(x-pi/4))+0.4 thank you so much!

asked by BreAnne on February 16, 2012
6. ### pre-cal

5 t values used to graph (called key pts or inflection pts) and primary cycle? 1) y=2sin(x-pi) 2) y=-sin3x-2 3) -3cos(1/2x+pi/3)+1 4) cospi(x-1)

asked by Kim on March 5, 2012
7. ### maths

Hello! I need help solving this equation. I need to find the value of t. Any help is greatly appreciated. 1.5=3cos(t) Thank you!

asked by castiel on September 9, 2014
8. ### Calculus Question! ASAP!

Hello! I have this problem: x(dx)/sqrt(9-x^2) I was wondering why I can't use trig substitution and substitute sqrt(9-x^2) for sqrt(1-sec^2) and having: integral x = 3sin(theta) dx = 3cos(theta)d(theata) integral

asked by Lucy on February 15, 2017
9. ### math (final trig problem)

Solve for x on the interval [0,π/2): cos^3(2x) + 3cos^2(2x) + 3cos(2x) = 4 I havent done trig for a while so what exactly does that mean in solving for x on that interval and how would i go about doing that? Where are you getting

asked by jacob on April 30, 2007
10. ### ap calc

find the maximum value of f(x)= 3cos(x)-2sin(x) I know you have to take the derivative but after the derivative I don't know how to solve it. f'(x)=-3sin(x)-2cos(x)=0 I don't know what to do after this please help.

asked by Anonymous on November 9, 2014

More Similar Questions