Find the linearization f(x)=1+x−−−−√ at x = 0, and then use it to approximate 0.96−−−−√ . (1 point) Responses L(x)=1−x2 , 1 L ( x ) = 1 − x 2 , 1 L(x)=x2−1 ], −12 L ( x ) = x 2 − 1 ], − 1 2 L(x)=1+x2 , 0.98 L ( x ) = 1 + x 2 , 0.98 L(x)=1+x2 , 1.05 L ( x ) = 1 + x 2 , 1.05 L(x)=1−x2 , 0.98
The linearization of \(f(x) = 1+\sqrt{x}\) at \(x = 0\) is given by:
\[L(x) = f(0) + f'(0)\cdot(x-0)\]
First, let's find \(f(0)\):
\[f(0) = 1 + \sqrt{0} = 1 + 0 = 1\]
Next, let's find \(f'(x)\):
\[f'(x) = \frac{d}{dx}\left(1+\sqrt{x}\right) = \frac{d}{dx}\left(1+x^{\frac{1}{2}}\right) = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}\]
Now, let's find \(f'(0)\):
\[f'(0) = \frac{1}{2\sqrt{0}} = \frac{1}{2\cdot 0} = \frac{1}{0}\]
Since the derivative is undefined at \(x = 0\), the linearization is not defined. Therefore, we cannot use it to approximate \(\sqrt{0.96}\).
The correct answer is: The linearization of \(f(x)=1+\sqrt{x}\) at \(x = 0\) is undefined and cannot be used to approximate \(\sqrt{0.96}\).