Let `f(x) = {{:((x^(3) + x^(2) -16x +20)/(x-2)^(2), If x ne 2),(=k, If x=2):}`, If `f(x)` is continous for all x, then k=

Options

A. 3
B. 5
C. 7
D. none

Answer

To find the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 2 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 2 equals \( f(2) \).

**Step 1: Define the function and the limit condition**

The function is defined as:
\[ f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2} & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases} \]
For \( f(x) \) to be continuous at \( x = 2 \), we need:
\[ \lim_{x \to 2} f(x) = f(2) = k \]

**Step 2: Calculate the limit as \( x \) approaches 2**

We need to compute:
\[ \lim_{x \to 2} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2} \]
First, we substitute \( x = 2 \) into the numerator:
\[ 2^3 + 2^2 - 16 \cdot 2 + 20 = 8 + 4 - 32 + 20 = 0 \]
The numerator evaluates to 0, and the denominator also evaluates to 0:
\[ (2-2)^2 = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

**Step 3: Apply L'Hôpital's Rule**

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit exists. Here, we differentiate the numerator and the denominator:

- The derivative of the numerator \( x^3 + x^2 - 16x + 20 \) is:
\[ 3x^2 + 2x - 16 \]

- The derivative of the denominator \( (x-2)^2 \) is:
\[ 2(x-2) \]

Thus, we have:
\[ \lim_{x \to 2} \frac{3x^2 + 2x - 16}{2(x-2)} \]

**Step 4: Substitute \( x = 2 \) again**

Substituting \( x = 2 \) into the new limit:
- The numerator becomes:
\[ 3(2^2) + 2(2) - 16 = 3(4) + 4 - 16 = 12 + 4 - 16 = 0 \]
- The denominator becomes:
\[ 2(2-2) = 2(0) = 0 \]
We still have the indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

**Step 5: Differentiate again**

Differentiate the numerator and denominator again:
- The second derivative of the numerator \( 3x^2 + 2x - 16 \) is:
\[ 6x + 2 \]

- The second derivative of the denominator \( 2(x-2) \) is:
\[ 2 \]

Now we compute the limit:
\[ \lim_{x \to 2} \frac{6x + 2}{2} \]
Substituting \( x = 2 \):
\[ \frac{6(2) + 2}{2} = \frac{12 + 2}{2} = \frac{14}{2} = 7 \]

**Step 6: Set the limit equal to \( k \)**

Since we found that:
\[ \lim_{x \to 2} f(x) = 7 \]
For continuity, we must have:
\[ k = 7 \]

Thus, the value of \( k \) is:
\[ \boxed{7} \]

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