Question

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!

Answer 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)

Answer 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 better be in radians
y x
0, 3
pi/6 , 7.26
-pi/6 , -.067
- pi/12 , +.54

so somewhere round -pi/6

Answer 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!

Answer 4

where did the 0 come from?
also you cannot have cos and cos in f'.