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

Question

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

Quizplus · Accepted Answer

The quadratic approximation of a function $f(x)$ at $x = a$ is given by $f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$. For $f(x) = \ln(\cos(3x))$ at $x = 0$, we first find $f(0) = \ln(\cos(0)) = \ln(1) = 0$, $f'(x) = -3\tan(3x)$ so $f'(0) = 0$, and $f''(x) = -9\sec^2(3x)$ so $f''(0) = -9$. Thus, the quadratic approximation is $0 + 0(x - 0) + \frac{-9}{2}(x - 0)^2 = -\frac{9}{2}x^2$.

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

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