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!
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)
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
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!
where did the 0 come from?
also you cannot have cos and cos in f'.
To find the inverse of a function, the general approach is to switch the roles of the independent variable (x) and dependent variable (f(x)) and solve for x. However, finding the inverse of a function can be challenging, especially if it contains trigonometric functions.
In this case, we are given f(x) = 7x + 3cos(x) + 2sin(x) and asked to compute (f^-1)(2). Let's break down the steps to find the inverse and then evaluate it for the given value.
Step 1: Express f(x) in terms of x and remove trigonometric functions if possible:
f(x) = 7x + 3cos(x) + 2sin(x)
Step 2: Switch the variables, replacing f(x) with y:
x = 7y + 3cos(y) + 2sin(y)
Step 3: Rearrange the equation to solve for y:
7y + 3cos(y) + 2sin(y) - x = 0
At this point, it becomes challenging to explicitly solve for y because of the combination of linear and trigonometric terms. This is where a numeric approximation or using computational methods becomes useful.
One approach is to use numerical methods, such as Newton's method or a graphing calculator, to approximate the solution for y.
Alternatively, if you suspect that the inverse is a piecewise function, you can attempt to solve for y within specific intervals. However, this approach can be more tedious and requires a good understanding of the behavior of the function.
Regarding the formulas you mentioned, (f^-1)(a) = (1/f'(f^-1(a)) is not a general approach for finding the inverse. It is specific to certain functions and requires knowledge of the derivative of the inverse function.
Therefore, without further analysis or using numerical approximations, it's not possible to determine the exact value of (f^-1)(2) or whether either of the given answers (1/2 or 1/10) is correct.
To summarize, finding the inverse of a function can be challenging, especially when it involves trigonometric functions. In such cases, numerical approximations or computational methods are often necessary.