To find the linear approximating polynomial for the function [tex]\( f(x) = \sqrt{9x + 9} \)[/tex] centered at [tex]\( a = 0 \)[/tex], follow these steps:
1. Evaluate [tex]\( f(a) \)[/tex]:
[tex]\[
f(0) = \sqrt{9(0) + 9} = \sqrt{9} = 3
\][/tex]
2. Find the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[
f(x) = \sqrt{9x + 9}
\][/tex]
To differentiate [tex]\( f(x) \)[/tex], we use the chain rule. Let [tex]\( u = 9x + 9 \)[/tex]. Then,
[tex]\[
f(x) = \sqrt{u} \Rightarrow f'(x) = \frac{d}{dx} (\sqrt{9x + 9}) = \frac{1}{2\sqrt{u}} \cdot \frac{d}{dx} (u)
\][/tex]
Since [tex]\( \frac{d}{dx} (u) = \frac{d}{dx} (9x + 9) = 9 \)[/tex], we substitute back:
[tex]\[
f'(x) = \frac{1}{2\sqrt{9x + 9}} \cdot 9 = \frac{9}{2\sqrt{9x + 9}}
\][/tex]
3. Evaluate [tex]\( f'(a) \)[/tex]:
[tex]\[
f'(0) = \frac{9}{2\sqrt{9 \cdot 0 + 9}} = \frac{9}{2\sqrt{9}} = \frac{9}{2 \cdot 3} = \frac{9}{6} = \frac{3}{2}
\][/tex]
4. Construct the linear approximation [tex]\( L(x) = f(a) + f'(a)(x - a) \)[/tex]:
[tex]\[
L(x) = f(0) + f'(0)(x - 0) = 3 + \frac{3}{2} x
\][/tex]
Thus, the linear approximating polynomial for the function centered at [tex]\( a = 0 \)[/tex] is:
[tex]\[
L(x) = \frac{3}{2} x + 3
\][/tex]
Therefore, the correct answer is:
D. [tex]\( L(x) = \frac{3}{2} x + 3 \)[/tex]
