Let `f(x)=-4.sqrt(e^(1-x))+1+x+(x^(2))/(2)+(x^(3))/(3)`. If g(x) is inverse of `f(x)` then the value of `(1)/(g^(')(-(7)/(6)))` is

Explanation

To solve the problem, we need to find the value of \( \frac{1}{g'(-\frac{7}{6})} \) where \( g(x) \) is the inverse of the function \( f(x) = -4\sqrt{e^{1-x}} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3} \).

### Step 1: Find \( f(1) \)

We first compute \( f(1) \):

\[ f(1) = -4\sqrt{e^{1-1}} + 1 + 1 + \frac{1^2}{2} + \frac{1^3}{3} \]

Calculating each term:

\[ f(1) = -4\sqrt{e^0} + 1 + 1 + \frac{1}{2} + \frac{1}{3} \]
\[ = -4(1) + 1 + 1 + \frac{1}{2} + \frac{1}{3} \]
\[ = -4 + 2 + \frac{1}{2} + \frac{1}{3} \]

Finding a common denominator (which is 6):

\[ = -4 + 2 + \frac{3}{6} + \frac{2}{6} = -4 + 2 + \frac{5}{6} \]
\[ = -2 + \frac{5}{6} = -\frac{12}{6} + \frac{5}{6} = -\frac{7}{6} \]

Thus, we have:

\[ f(1) = -\frac{7}{6} \]

### Step 2: Use the Inverse Function Theorem

From the inverse function theorem, we know that:

\[ g'(f(x)) \cdot f'(x) = 1 \]

Thus, we can express \( g'(-\frac{7}{6}) \) in terms of \( f'(1) \):

\[ g'(-\frac{7}{6}) = \frac{1}{f'(1)} \]

### Step 3: Differentiate \( f(x) \)

Now we need to find \( f'(x) \):

\[ f'(x) = \frac{d}{dx}\left(-4\sqrt{e^{1-x}} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}\right) \]

Using the chain rule for the first term:

\[ f'(x) = -4 \cdot \frac{1}{2\sqrt{e^{1-x}}} \cdot (-e^{1-x}) + 1 + x + x^2 \]
\[ = \frac{4e^{1-x}}{2\sqrt{e^{1-x}}} + 1 + x + x^2 \]
\[ = \frac{2e^{1-x}}{\sqrt{e^{1-x}}} + 1 + x + x^2 \]
\[ = 2\sqrt{e^{1-x}} + 1 + x + x^2 \]

### Step 4: Evaluate \( f'(1) \)

Now we evaluate \( f'(1) \):

\[ f'(1) = 2\sqrt{e^{1-1}} + 1 + 1 + 1^2 \]
\[ = 2\sqrt{e^0} + 1 + 1 + 1 = 2(1) + 1 + 1 + 1 = 5 \]

### Step 5: Find \( g'(-\frac{7}{6}) \)

Now substituting back into our equation for \( g'(-\frac{7}{6}) \):

\[ g'(-\frac{7}{6}) = \frac{1}{f'(1)} = \frac{1}{5} \]

### Step 6: Find \( \frac{1}{g'(-\frac{7}{6})} \)

Finally, we find:

\[ \frac{1}{g'(-\frac{7}{6})} = 5 \]

Thus, the final answer is:

\[ \boxed{5} \]