Find the quadratic approximation of f at x = 0.
-f(x) = ln(1 + sin 3x)
A) $Q ( x ) = 1 - 3 x + \frac { 9 } { 2 } x ^ { 2 }$
B) $Q ( x ) = 3 x + \frac { 9 } { 2 } x ^ { 2 }$
C) $Q ( x ) = 1 + 3 x + \frac { 9 } { 2 } x ^ { 2 }$
D) $Q ( x ) = 3 x - \frac { 9 } { 2 } x ^ { 2 }$

Question

Find the quadratic approximation of f at x = 0.
-f(x) = ln(1 + sin 3x) 
A) $Q ( x ) = 1 - 3 x + \frac { 9 } { 2 } x ^ { 2 }$
B) $Q ( x ) = 3 x + \frac { 9 } { 2 } x ^ { 2 }$
C) $Q ( x ) = 1 + 3 x + \frac { 9 } { 2 } x ^ { 2 }$
D) $Q ( x ) = 3 x - \frac { 9 } { 2 } x ^ { 2 }$

Quizplus · Accepted Answer

The quadratic approximation of a function $f(x)$ at $x = a$ is given by $Q(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$. For $f(x) = \ln(1 + \sin(3x))$ at $x = 0$:1. $f(0) = \ln(1 + \sin(0)) = \ln(1) = 0$.2. $f'(x) = \frac{1}{1 + \sin(3x)} \cdot \cos(3x) \cdot 3$, so $f'(0) = \frac{3}{1} \cdot 1 = 3$.3. $f''(x)$ involves differentiating $f'(x)$ again, which results in $f''(0) = -9$.Therefore, $Q(x) = 0 + 3x + \frac{-9}{2}x^2 = 3x - \frac{9}{2}x^2$, which matches option D.

Find the Quadratic Approximation of F at X = 0 Q(x)=1−3x+92x2Q ( x ) = 1 - 3 x + \frac { 9 } { 2 } x ^ { 2 }Q(x)=1−3x+29​x2

Find the Quadratic Approximation of F at X = 0 $Q ( x ) = 1 - 3 x + \frac { 9 } { 2 } x ^ { 2 }$