If `f(x)={(2^(x+2)-16)/(4^x-16),if x!=2k ,if f(x) is continuous a t x=2, find k`

Explanation

To solve the problem, we need to find the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{2^{x+2} - 16}{4^x - 16} & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases} \]

is continuous at \( x = 2 \).

### Step 1: Determine the limit of \( f(x) \) as \( x \) approaches 2.

To ensure continuity at \( x = 2 \), we need:

\[ \lim_{x \to 2} f(x) = f(2) = k \]

We will calculate \( \lim_{x \to 2} f(x) \):

\[ \lim_{x \to 2} \frac{2^{x+2} - 16}{4^x - 16} \]

### Step 2: Simplify the expression.

First, we can rewrite \( 4^x \) as \( (2^2)^x = 2^{2x} \):

\[ \lim_{x \to 2} \frac{2^{x+2} - 16}{2^{2x} - 16} \]

Now, substituting \( 16 \) as \( 2^4 \):

\[ = \lim_{x \to 2} \frac{2^{x+2} - 2^4}{2^{2x} - 2^4} \]

### Step 3: Factor the numerator and denominator.

Both the numerator and denominator can be factored using the difference of squares:

1. The numerator \( 2^{x+2} - 2^4 = 2^{x+2} - 16 \) can be factored as:

\[ 2^{x+2} - 16 = (2^{x+2} - 4)(2^{x+2} + 4) \]

2. The denominator \( 2^{2x} - 16 = 2^{2x} - 4^2 \) can be factored as:

\[ 2^{2x} - 16 = (2^x - 4)(2^x + 4) \]

### Step 4: Substitute \( x = 2 \).

Now we can substitute \( x = 2 \) into the limit:

\[ = \lim_{x \to 2} \frac{(2^{x+2} - 4)(2^{x+2} + 4)}{(2^x - 4)(2^x + 4)} \]

### Step 5: Evaluate the limit.

At \( x = 2 \):

- \( 2^{x+2} = 2^4 = 16 \)
- \( 2^x = 2^2 = 4 \)

So we have:

\[ = \frac{(16 - 4)(16 + 4)}{(4 - 4)(4 + 4)} = \frac{(12)(20)}{(0)(8)} \]

This indicates that we have a \( 0/0 \) form, so we can apply L'Hôpital's Rule or simplify further.

### Step 6: Cancel common factors.

Notice that \( 2^x - 4 \) is a factor in both the numerator and denominator. We can simplify:

\[ = \lim_{x \to 2} \frac{(2^{x+2} - 16)}{(2^x - 4)} \cdot \frac{1}{(2^x + 4)} \]

### Step 7: Find the limit.

Now we can differentiate the numerator and denominator or substitute values directly after canceling:

\[ = \lim_{x \to 2} \frac{(2^{x+2} - 16)}{(2^x - 4)} \cdot \frac{1}{(2^x + 4)} \]

After simplification, we find:

\[ = \frac{12}{8} = \frac{3}{2} \]

### Step 8: Set the limit equal to \( k \).

Thus, we have:

\[ k = \frac{3}{2} \]

### Final Answer:

The value of \( k \) is \( \frac{3}{2} \).

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