To find the linearization [tex]\(L(x)\)[/tex] of the function [tex]\(f(x) = \sqrt{x^2 + 27}\)[/tex] at [tex]\(x = -3\)[/tex], we follow these steps:
1. Evaluate the function at [tex]\(x = -3\)[/tex]:
[tex]\[
f(-3) = \sqrt{(-3)^2 + 27} = \sqrt{9 + 27} = \sqrt{36} = 6
\][/tex]
2. Find the derivative [tex]\(f'(x)\)[/tex]:
[tex]\[
f(x) = \sqrt{x^2 + 27}
\][/tex]
Using the chain rule to differentiate:
[tex]\[
f'(x) = \frac{d}{dx} \left( x^2 + 27 \right)^{1/2} = \frac{1}{2} (x^2 + 27)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 27}}
\][/tex]
3. Evaluate the derivative at [tex]\(x = -3\)[/tex]:
[tex]\[
f'(-3) = \frac{-3}{\sqrt{(-3)^2 + 27}} = \frac{-3}{\sqrt{36}} = \frac{-3}{6} = -\frac{1}{2}
\][/tex]
4. Form the linearization [tex]\(L(x)\)[/tex] using the point-slope form of the tangent line:
The linearization of [tex]\(f(x)\)[/tex] at [tex]\(x = a\)[/tex] is given by:
[tex]\[
L(x) = f(a) + f'(a)(x - a)
\][/tex]
For our function at [tex]\(x = -3\)[/tex]:
[tex]\[
L(x) = f(-3) + f'(-3)(x - (-3)) = 6 + \left( -\frac{1}{2} \right)(x + 3)
\][/tex]
5. Simplify the expression:
[tex]\[
L(x) = 6 - \frac{1}{2}(x + 3) = 6 - \frac{1}{2}x - \frac{3}{2} = \frac{12}{2} - \frac{3}{2} - \frac{1}{2}x = \frac{9}{2} - \frac{1}{2}x
\][/tex]
Thus, the linearization [tex]\(L(x)\)[/tex] of [tex]\(f(x)\)[/tex] at [tex]\(x = -3\)[/tex] is:
[tex]\[
L(x) = \frac{9}{2} - \frac{1}{2}x
\][/tex]
