Question

use linear approximation, i.e. the tangent line, to approximate $\sqrt{49.2}$ as follows:
let $f(x)=\sqrt{x}$. find the equation of the tangent line to $f(x)$ at $x = 49$
$l(x)=$
using this, we find our approximation for $\sqrt{49.2}$ is
note: for this part, give your answer to at least 9 significant figures or use an expression to give the exact answer.
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Explanation:

Step1: Find the derivative of \(f(x)\)

The function \(f(x)=\sqrt{x}=x^{\frac{1}{2}}\). Using the power rule \((x^n)^\prime = nx^{n - 1}\), we have \(f^\prime(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).
When \(x = 49\), \(f(49)=\sqrt{49}=7\) and \(f^\prime(49)=\frac{1}{2\sqrt{49}}=\frac{1}{14}\).

Step2: Use the point - slope form of a line

The point - slope form of a line is \(y - y_0=f^\prime(x_0)(x - x_0)\). Here \(x_0 = 49\) and \(y_0=f(49) = 7\).
So \(L(x)-7=\frac{1}{14}(x - 49)\), then \(L(x)=\frac{1}{14}(x - 49)+7=\frac{1}{14}x-\frac{49}{14}+7=\frac{1}{14}x-\frac{7}{2}+7=\frac{1}{14}x+\frac{7}{2}\).

Step3: Approximate \(\sqrt{49.2}\)

We want to approximate \(f(49.2)\) using \(L(x)\). Substitute \(x = 49.2\) into \(L(x)\).
\(L(49.2)=\frac{1}{14}\times49.2+\frac{7}{2}=\frac{49.2}{14}+3.5\).
\(\frac{49.2}{14}=3.514285714\), and \(3.514285714+3.5 = 7.014285714\).

Answer:

\(L(x)=\frac{1}{14}x+\frac{7}{2}\); the approximation for \(\sqrt{49.2}\) is \(7.014285714\)