If g is inverse of `f(x) = x^3 + x + cosx`, then find the value of `g'(1)`

Explanation

To find the value of \( g'(1) \) where \( g \) is the inverse of the function \( f(x) = x^3 + x + \cos x \), we can use the formula for the derivative of the inverse function. The formula states that if \( g \) is the inverse of \( f \), then:

\[ g'(y) = \frac{1}{f'(g(y))} \]

In this case, we need to find \( g'(1) \). This means we first need to find \( x \) such that \( f(x) = 1 \).

### Step 1: Solve for \( x \) such that \( f(x) = 1 \)

We need to solve the equation:

\[ f(x) = x^3 + x + \cos x = 1 \]

Let's check \( x = 0 \):

\[ f(0) = 0^3 + 0 + \cos(0) = 0 + 0 + 1 = 1 \]

So, \( f(0) = 1 \). This means \( g(1) = 0 \).

### Step 2: Find \( f'(x) \)

Next, we need to find the derivative \( f'(x) \):

\[ f'(x) = \frac{d}{dx}(x^3 + x + \cos x) = 3x^2 + 1 - \sin x \]

### Step 3: Evaluate \( f'(0) \)

Now we evaluate \( f'(0) \):

\[ f'(0) = 3(0)^2 + 1 - \sin(0) = 0 + 1 - 0 = 1 \]

### Step 4: Use the inverse function derivative formula

Now we can use the formula for the derivative of the inverse function:

\[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)} = \frac{1}{1} = 1 \]

### Conclusion

Thus, the value of \( g'(1) \) is:

\[ \boxed{1} \]