If the function `f(x) ={:{((3x^3-2x^2-1)/(x-1)", "x ne 1),(" "K", " x= 1):},`
is continuous at x=1,find the value of k.

Answer

To find the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{3x^3 - 2x^2 - 1}{x - 1} & \text{if } x \neq 1 \\ k & \text{if } x = 1 \end{cases} \]

is continuous at \( x = 1 \), we need to ensure that

\[ f(1) = \lim_{x \to 1} f(x). \]

### Step 1: Set up the equation for continuity
Since \( f(1) = k \), we need to find

\[ k = \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{3x^3 - 2x^2 - 1}{x - 1}. \]

### Step 2: Evaluate the limit
Substituting \( x = 1 \) directly into the limit gives us

\[ \frac{3(1)^3 - 2(1)^2 - 1}{1 - 1} = \frac{3 - 2 - 1}{0} = \frac{0}{0}. \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator:

1. Differentiate the numerator:
\[ \frac{d}{dx}(3x^3 - 2x^2 - 1) = 9x^2 - 4x. \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(x - 1) = 1. \]

Now we can rewrite the limit:

\[ \lim_{x \to 1} \frac{3x^3 - 2x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{9x^2 - 4x}{1}. \]

### Step 4: Substitute \( x = 1 \) into the new limit
Now we substitute \( x = 1 \):

\[ 9(1)^2 - 4(1) = 9 - 4 = 5. \]

### Step 5: Conclude the value of \( k \)
Thus, we have

\[ k = 5. \]

### Final Answer
The value of \( k \) is \( 5 \).

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