find the tangent line approximation for $\\sqrt{9 + x}$ near $x = 0$.\n$y =$

Explanation:

Step1: Recall the formula for tangent - line approximation

The tangent - line approximation (linear approximation) of a function (y = f(x)) at (x = a) is given by (L(x)=f(a)+f^{\prime}(a)(x - a)). Let (f(x)=\sqrt{9 + x}=(9 + x)^{\frac{1}{2}}), and (a = 0).

Step2: Find (f(a))

First, find (f(0)). Substitute (x = 0) into (f(x)): (f(0)=\sqrt{9+0}=3).

Step3: Find the derivative of (f(x))

Using the power rule ((u^{n})^\prime=nu^{n - 1}u^\prime), where (u = 9 + x), (n=\frac{1}{2}), and (u^\prime=1). (f^{\prime}(x)=\frac{1}{2}(9 + x)^{-\frac{1}{2}}\times(9 + x)^\prime=\frac{1}{2\sqrt{9 + x}}).

Step4: Find (f^{\prime}(a))

Substitute (x = 0) into (f^{\prime}(x)): (f^{\prime}(0)=\frac{1}{2\sqrt{9+0}}=\frac{1}{6}).

Step5: Use the linear - approximation formula

Substitute (a = 0), (f(a)=3), and (f^{\prime}(a)=\frac{1}{6}) into (L(x)=f(a)+f^{\prime}(a)(x - a)). (L(x)=3+\frac{1}{6}(x - 0)).

Answer:

(y = 3+\frac{1}{6}x)