The linear approximation of f(x)=sqrt (x+1) at a=3 is f(x)=(1)/(4)x-(5)/(4) f(x)=(1)/(2)x-(1)/(4) f(x)=(1)/(4)x+(5)/(4) f(x)=(1)/(2)x-(1)/(2) f(x)=(1)/(2)x+(1)/(2)

## Step 1: Calculate $f(a)$<br />### We are given $f(x) = \sqrt{x+1}$ and $a=3$. Therefore, $f(a) = f(3) = \sqrt{3+1} = \sqrt{4} = 2$.<br /><br />## Step 2: Calculate $f'(x)$<br />### The derivative of $f(x) = \sqrt{x+1} = (x+1)^{\frac{1}{2}}$ is $f'(x) = \frac{1}{2}(x+1)^{-\frac{1}{2}} = \frac{1}{2\sqrt{x+1}}$.<br /><br />## Step 3: Calculate $f'(a)$<br />### Substituting $a=3$ into $f'(x)$, we get $f'(a) = f'(3) = \frac{1}{2\sqrt{3+1}} = \frac{1}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4}$.<br /><br />## Step 4: Construct the linear approximation<br />### The linear approximation of $f(x)$ at $a$ is given by $L(x) = f(a) + f'(a)(x-a)$. Substituting the values we calculated, we get $L(x) = 2 + \frac{1}{4}(x-3) = 2 + \frac{1}{4}x - \frac{3}{4} = \frac{1}{4}x + \frac{8-3}{4} = \frac{1}{4}x + \frac{5}{4}$.