Find the linear approximation for $$ f(x)=\sqrt{x+3} $$ at $$ a=1 $$

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To find the **linear approximation** of the function $$ f(x) = \sqrt{x + 3} $$ at $$ a = 1 $$, we'll use the formula for the linear (tangent line) approximation:

$$
L(x) = f(a) + f'(a)(x - a)
$$

**Step 1: Compute $$ f(a) $$**

$$
f(1) = \sqrt{1 + 3} = \sqrt{4} = 2
$$

**Step 2: Find the derivative $$ f'(x) $$**

$$
f(x) = (x + 3)^{1/2} \
f'(x) = \frac{1}{2}(x + 3)^{-1/2} = \frac{1}{2\sqrt{x + 3}}
$$

**Step 3: Evaluate $$ f'(a) $$ at $$ a = 1 $$**

$$
f'(1) = \frac{1}{2\sqrt{1 + 3}} = \frac{1}{2 \times 2} = \frac{1}{4}
$$

**Step 4: Construct the linear approximation $$ L(x) $$**

$$
L(x) = f(1) + f'(1)(x - 1) = 2 + \frac{1}{4}(x - 1)
$$

**Final Answer:**
$$
L(x) = 2 + \frac{1}{4}(x - 1)
$$

Which can also be written as:

$$
L(x) = \frac{1}{4}x + \frac{7}{4}
$$

**Example Usage:**
If you want to approximate $$ f(1.2) $$:

$$
L(1.2) = 2 + \frac{1}{4}(1.2 - 1) = 2 + \frac{1}{4}(0.2) = 2 + 0.05 = 2.05
$$

This linear approximation provides a good estimate of the function near $$ x = 1 $$.

**Final Answer:**
A linear approximation is 2 plus one fourth of (x – 1). In symbols,

L(x) = 2 + ¼ (x − 1)
A linear approximation of $$ f(x) = \sqrt{x + 3} $$ at $$ a = 1 $$ is: $$ L(x) = 2 + \frac{1}{4}(x - 1) $$

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