The function `f(x)=e^(x^(3)-6x^(2)+10)` attains local extremum at x = a and x = b (a < b), then the value of `a+b` is equal to

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) where the function \( f(x) = e^{x^3 - 6x^2 + 10} \) attains local extremum, and then calculate \( a + b \).

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( e^{x^3 - 6x^2 + 10} \right) \]
Using the chain rule, we have:
\[ f'(x) = e^{x^3 - 6x^2 + 10} \cdot \frac{d}{dx}(x^3 - 6x^2 + 10) \]
Next, we differentiate the exponent:
\[ \frac{d}{dx}(x^3 - 6x^2 + 10) = 3x^2 - 12x \]
Thus, the derivative becomes:
\[ f'(x) = e^{x^3 - 6x^2 + 10} \cdot (3x^2 - 12x) \]

### Step 2: Set the derivative to zero
To find the critical points where the function attains local extremum, we set the derivative equal to zero:
\[ e^{x^3 - 6x^2 + 10} \cdot (3x^2 - 12x) = 0 \]
Since \( e^{x^3 - 6x^2 + 10} \) is never zero, we can focus on the quadratic part:
\[ 3x^2 - 12x = 0 \]

### Step 3: Factor the equation
We can factor out the common term:
\[ 3x(x - 4) = 0 \]
Setting each factor to zero gives us:
\[ 3x = 0 \quad \Rightarrow \quad x = 0 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

### Step 4: Identify the values of \( a \) and \( b \)
From the above calculations, we have found the critical points:
\[ x = 0 \quad \text{and} \quad x = 4 \]
Given that \( a < b \), we can assign:
\[ a = 0 \quad \text{and} \quad b = 4 \]

### Step 5: Calculate \( a + b \)
Now, we can find \( a + b \):
\[ a + b = 0 + 4 = 4 \]

### Final Answer
Thus, the value of \( a + b \) is:
\[ \boxed{4} \]