Question

find the local linear approximation of the function ( f(x)=sqrt{1 + x} ) at ( x_0 = 3 ), and use it to approximate ( sqrt{3.9} ) and ( sqrt{4.1} ).
(a) ( f(x)=sqrt{1 + x}approx )
(b) ( sqrt{3.9}approx )
(c) ( sqrt{4.1}approx )
note: you can earn partial credit on this problem.
for parts (b) and (c), you should enter your answer as a fraction. if you enter a decimal, make sure that it is correct to at least six decimal places.

Explanation:

Step1: Recall the formula for local linear approximation

The formula for the local linear approximation of a function \(y = f(x)\) at \(x = x_0\) is \(L(x)=f(x_0)+f^{\prime}(x_0)(x - x_0)\). Given \(f(x)=\sqrt{1 + x}=(1 + x)^{\frac{1}{2}}\), we first find its derivative. Using the power - rule \((u^n)^\prime=nu^{n - 1}u^\prime\), if \(u = 1+x\) and \(n=\frac{1}{2}\), then \(f^{\prime}(x)=\frac{1}{2}(1 + x)^{-\frac{1}{2}}\).
When \(x_0 = 3\), \(f(3)=\sqrt{1+3}=2\), and \(f^{\prime}(3)=\frac{1}{2}(1 + 3)^{-\frac{1}{2}}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\). So the local linear approximation \(L(x)=2+\frac{1}{4}(x - 3)=\frac{1}{4}x+\frac{8 - 3}{4}=\frac{1}{4}x+\frac{5}{4}\).

Step2: Approximate \(\sqrt{3.9}\)

Let \(x = 3.9\). Substitute \(x = 3.9\) into \(L(x)\): \(L(3.9)=\frac{1}{4}\times3.9+\frac{5}{4}=\frac{3.9 + 5}{4}=\frac{8.9}{4}=\frac{89}{40}\).

Step3: Approximate \(\sqrt{4.1}\)

Let \(x = 4.1\). Substitute \(x = 4.1\) into \(L(x)\): \(L(4.1)=\frac{1}{4}\times4.1+\frac{5}{4}=\frac{4.1+5}{4}=\frac{9.1}{4}=\frac{91}{40}\).

Answer:

(a) \(\frac{89}{40}\) (b) \(\frac{91}{40}\)