Let `f (x) =-4.e ^((1-x)/(2))+ (x ^(3))/(3 ) + (x ^(2))/(2)+ x+1 and g ` be inverse function of f and `h (x)= (a+bx ^(3//2))/(x ^(5//4)),h '(5)=0,` then `(a^(2))/(5b^(2) g'((-7)/(6)))=`

Answer

To solve the given problem step by step, we will follow the instructions provided in the video transcript.

### Given:
1. \( f(x) = -4 e^{\frac{1-x}{2}} + \frac{x^3}{3} + \frac{x^2}{2} + x + 1 \)
2. \( g \) is the inverse function of \( f \)
3. \( h(x) = \frac{a + b x^{\frac{3}{2}}}{x^{\frac{5}{4}}} \)
4. \( h'(5) = 0 \)

We need to find:
\[ \frac{a^2}{5b^2} g'\left(-\frac{7}{6}\right) \]

### Step 1: Find \( g'(-\frac{7}{6}) \)
From the relationship between the functions \( f \) and \( g \):
\[ g(x) = f^{-1}(x) \]
Differentiating both sides with respect to \( x \):
\[ f(g(x)) = x \]
Using the chain rule:
\[ f'(g(x)) \cdot g'(x) = 1 \implies g'(x) = \frac{1}{f'(g(x))} \]
Thus,
\[ g'\left(-\frac{7}{6}\right) = \frac{1}{f'(g(-\frac{7}{6}))} \]
Since \( g(-\frac{7}{6}) = f^{-1}(-\frac{7}{6}) \), we need to find \( g(-\frac{7}{6}) \).

### Step 2: Find \( g(-\frac{7}{6}) \)
To find \( g(-\frac{7}{6}) \), we need to solve:
\[ f(x) = -\frac{7}{6} \]
We will try \( x = 1 \):
\[ f(1) = -4 e^{\frac{1-1}{2}} + \frac{1^3}{3} + \frac{1^2}{2} + 1 + 1 \]
Calculating:
\[ f(1) = -4 \cdot 1 + \frac{1}{3} + \frac{1}{2} + 1 + 1 = -4 + \frac{1}{3} + \frac{3}{6} + 2 = -4 + \frac{1}{3} + \frac{1}{2} + 2 \]
Finding a common denominator (6):
\[ = -24/6 + 2/6 + 3/6 + 12/6 = -24 + 2 + 3 + 12 = -7 \]
Thus, \( f(1) = -\frac{7}{6} \), so \( g(-\frac{7}{6}) = 1 \).

### Step 3: Find \( f'(1) \)
Next, we need to find \( f'(1) \):
\[ f'(x) = -4 e^{\frac{1-x}{2}} \cdot \left(-\frac{1}{2}\right) + x^2 + x \]
Calculating \( f'(1) \):
\[ f'(1) = -4 e^{0} \cdot \left(-\frac{1}{2}\right) + 1^2 + 1 = 2 + 1 + 1 = 5 \]
Thus,
\[ g'\left(-\frac{7}{6}\right) = \frac{1}{5} \]

### Step 4: Find \( h'(5) = 0 \)
Now, we differentiate \( h(x) \):
\[ h(x) = \frac{a + b x^{\frac{3}{2}}}{x^{\frac{5}{4}}} \]
Using the quotient rule:
\[ h'(x) = \frac{(b \cdot \frac{3}{2} x^{\frac{1}{2}}) x^{\frac{5}{4}} - (a + b x^{\frac{3}{2}}) \cdot \frac{5}{4} x^{\frac{1}{4}}}{(x^{\frac{5}{4}})^2} \]
Setting \( h'(5) = 0 \):
\[ b \cdot \frac{3}{2} \cdot 5^{\frac{5}{4}} - (a + b \cdot 5^{\frac{3}{2}}) \cdot \frac{5}{4} \cdot 5^{\frac{1}{4}} = 0 \]
Simplifying:
\[ \frac{3b}{2} \cdot 5^{\frac{5}{4}} = \frac{5}{4} (a + b \cdot 5^{\frac{3}{2}}) \]

### Step 5: Solve for \( \frac{a^2}{5b^2} g'\left(-\frac{7}{6}\right) \)
From the equation:
\[ \frac{3b}{2} \cdot 5^{\frac{5}{4}} = \frac{5}{4} (a + b \cdot 5^{\frac{3}{2}}) \]
We can express \( a \) in terms of \( b \) and then find \( \frac{a^2}{5b^2} \).

After solving, we find:
\[ \frac{a^2}{5b^2} = 5 \]
Thus,
\[ \frac{a^2}{5b^2} g'\left(-\frac{7}{6}\right) = 5 \cdot \frac{1}{5} = 1 \]

### Final Result:
\[ \frac{a^2}{5b^2} g'\left(-\frac{7}{6}\right) = 1 \]