Find 'a' if `f^(')(a)=0`, where `f(x)=x^(3)-3x^(2)+3x-1`

Answer

To find the value of 'a' such that \( f'(a) = 0 \) for the function \( f(x) = x^3 - 3x^2 + 3x - 1 \), we will follow these steps:

### Step 1: Differentiate the function \( f(x) \)
We need to find the derivative \( f'(x) \).

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

Using the power rule of differentiation:

\[ f'(x) = 3x^2 - 6x + 3 \]

### Step 2: Set the derivative equal to zero
Now, we need to find the values of \( x \) for which \( f'(x) = 0 \):

\[ 3x^2 - 6x + 3 = 0 \]

### Step 3: Simplify the equation
We can simplify this equation by dividing all terms by 3:

\[ x^2 - 2x + 1 = 0 \]

### Step 4: Factor the quadratic equation
Now we can factor the quadratic:

\[ (x - 1)^2 = 0 \]

### Step 5: Solve for \( x \)
Setting the factor equal to zero gives us:

\[ x - 1 = 0 \implies x = 1 \]

### Conclusion
Thus, the value of \( a \) is:

\[ a = 1 \]