Let $$ f(x)=\sqrt{x+1} $$ and use the Linear Approximation to this function at $$ a=3 $$ with $$ \Delta x=0.5 $$ to estimate $$ f(3.5)-f(3)=\Delta f \approx d f=\square $$ Preview My Answers Submit Answers

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**Step 1. Compute $$ f(3) $$:**  
$$
f(3) = \sqrt{3+1} = \sqrt{4} = 2.
$$

**Step 2. Compute the derivative $$ f'(x) $$:**  
Given $$ f(x)=\sqrt{x+1} $$, rewrite it as:  
$$
f(x) = (x+1)^{\frac{1}{2}}.
$$  
Differentiate with respect to $$ x $$:  
$$
f'(x) = \frac{1}{2}(x+1)^{-\frac{1}{2}}.
$$

**Step 3. Evaluate $$ f'(x) $$ at $$ x = 3 $$:**  
$$
f'(3) = \frac{1}{2}(3+1)^{-\frac{1}{2}} = \frac{1}{2}(4)^{-\frac{1}{2}} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.
$$

**Step 4. Use linear approximation to estimate $$ \Delta f $$:**  
For a small change $$ \Delta x $$,  
$$
\Delta f \approx df = f'(3) \cdot \Delta x.
$$  
Given $$ \Delta x = 0.5 $$:  
$$
df = \frac{1}{4} \cdot 0.5 = \frac{1}{8}.
$$

**Final Answer:**  
$$
\text{df} = \frac{1}{8}.
$$
$$ \Delta f \approx \frac{1}{8} $$

Let \( f(x)=\sqrt{x+1} \) and use the Linear Approximation to this function at \( a=3 \) with \( \Delta x=0.5 \) to estimate \( f(3.5)-f(3)=\Delta f \approx d f=\square \) Preview My Answers Submit Answers

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