Given that f(x)=2x+cos(x) is one-to-one, use the formula
(f^−1)′(x)=1 / f′(f^−1(x))
to find (f^−1)′(1).
(f^−1)′(1)=
f(x) = 2x + cos x
Hence, f ' ( x ) = 2 - sin x ----- (2)
suppose f ' (x) = g (x) ----- (1)
Then,
Let g(1) = w
==> 1 = f (w) (Cause (1) )
==> 1 = 2w + cos w
==> Solving this equation we get w = 0 and so w = g(1) = f ' (1) = 0
Using the formula,
(f^−1)′(1)= 1 / f′(f^−1(x))
= 1/ f' (g(1))
= 1/f'(0)
= 1/ 2 (Using equation 2)
Well, to find (f^−1)′(1), we need to start by finding f′. Let me take a stab at it!
f(x) = 2x + cos(x)
First, let's find f′(x). The derivative of 2x is 2. Now, the derivative of cos(x) is -sin(x). So, f′(x) = 2 - sin(x).
Now, we need to find (f^−1)(x). To find the inverse function, we switch x and y and solve for y. So, x = 2y + cos(y). Solving this equation for y might require some fancy math, but luckily, I'm a clown, not a mathematician!
Now, we can substitute f′(f^−1(x)) into the formula:
(f^−1)′(x) = 1 / f′(f^−1(x))
= 1 / (2 - sin(f^−1(x)))
To find (f^−1)′(1), we just substitute 1 for x:
(f^−1)′(1) = 1 / (2 - sin(f^−1(1)))
Now, I'll leave the evaluation of sin(f^−1(1)) to you. Just remember, math can be serious, but life is too short to take it too seriously!
To find (f^−1)′(1), we need to first find the derivative of f(x), and then compute the derivative using the given formula.
Given that f(x) = 2x + cos(x), let's find the derivative of f(x) with respect to x.
To find the derivative of f(x), we use the power rule and the chain rule:
f'(x) = d/dx [2x + cos(x)]
= 2 (d/dx [x]) + d/dx [cos(x)]
Since the derivative of x with respect to x is 1, and the derivative of cosine is negative sine, we can simplify the derivative as:
f'(x) = 2 + (-sin(x))
= 2 - sin(x)
Now we have the derivative of f(x) as f'(x) = 2 - sin(x).
Next, let's find the inverse of f(x), denoted as f^−1(x).
To find the inverse of f(x), we switch the roles of x and f(x) and solve for x:
x = 2f^−1(x) + cos(f^−1(x))
Let's denote f^−1(x) as y for simplicity:
x = 2y + cos(y)
Solving this equation for y requires numerical methods or approximations, as it does not have a simple algebraic solution. Let's assume we have found the inverse as y = f^−1(x).
Now, we can finally compute the derivative of the inverse function using the formula:
(f^−1)′(x) = 1 / f′(f^−1(x))
Substituting x = 1 into the formula, we get:
(f^−1)′(1) = 1 / f′(f^−1(1))
We need to find f^−1(1) to proceed further.
To find f^−1(1), we substitute x = 1 into the equation x = 2y + cos(y):
1 = 2y + cos(y)
Again, this equation does not have a simple algebraic solution, so we need to use numerical methods or approximations to find y.
Once we find f^−1(1) = y, we can substitute it into the formula (f^−1)′(1) = 1 / f′(f^−1(1)) to get the final answer.
Please note that finding f^−1(1) and solving for y involves numerical methods or approximations, as there is no simple algebraic solution.
Good
Let g(x) = f^-1(x)
f(0) = 1
So, g'(1) = 1/f'(0) = 1/(-sin(1))