Find the quadratic approximation of f at x = 0.
-$$f ( x ) = e ^ { \sin 6 x }$$
A) $Q ( x ) = 6 x + 18 x ^ { 2 }$
B) $Q ( x ) = 6 x - 18 x ^ { 2 }$
C) $Q ( x ) = 1 - 6 x + 18 x ^ { 2 }$
D) $Q ( x ) = 1 + 6 x + 18 x ^ { 2 }$

Question

Quizplus · Accepted Answer

To find the quadratic approximation of $f(x) = e^{\sin 6x}$ at $x = 0$, we use the formula for the quadratic approximation, which is $f(x) \approx f(0) + f'(0)x + \frac{1}{2}f''(0)x^2$.First, calculate $f(0)$, $f'(x)$, and $f''(x)$:1. $f(0) = e^{\sin(0)} = e^0 = 1$.2. To find $f'(x)$, we use the chain rule: $f'(x) = e^{\sin 6x} \cdot \cos 6x \cdot 6 = 6e^{\sin 6x} \cos 6x$. Thus, $f'(0) = 6e^{\sin(0)} \cos(0) = 6$.3. For $f''(x)$, we differentiate $f'(x)$ again: $f''(x) = 6 \cdot [\cos 6x \cdot 6e^{\sin 6x} - \sin 6x \cdot 6e^{\sin 6x}]$. Therefore, $f''(0) = 6 \cdot [6 - 0] = 36$.Using these values in the quadratic approximation formula, we get:$$Q(x) = 1 + 6x + \frac{1}{2}(36)x^2 = 1 + 6x + 18x^2$$.Therefore, the correct answer is $Q(x) = 1 + 6x + 18x^2$.

Find the Quadratic Approximation of F at X = 0 f(x)=esin⁡6xf ( x ) = e ^ { \sin 6 x }f(x)=esin6x

Find the Quadratic Approximation of F at X = 0 $f ( x ) = e ^ { \sin 6 x }$