Question

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

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:

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