Find the quadratic approximation of f at x = 0.
-$$f ( x ) = \frac { x } { \sqrt { 81 - x ^ { 2 } } }$$
A) $Q ( x ) = \frac { x } { 9 }$
B) $Q ( x ) = 1 + 9 x$
C) $Q ( x ) = 1 + \frac { x } { 9 }$
D) $Q ( x ) = 9 x$

Question

Quizplus · Accepted Answer

The quadratic approximation of a function $f(x)$ at a point $x = a$ is given by $Q(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$. For $f(x) = \frac{x}{\sqrt{81 - x^2}}$ at $x = 0$, we first find $f(0)$, $f'(x)$, and $f''(x)$.1. $f(0) = \frac{0}{\sqrt{81 - 0^2}} = 0$.2. To find $f'(x)$, we use the quotient rule and chain rule:$$f'(x) = \frac{\sqrt{81 - x^2} - x \cdot \frac{-2x}{2\sqrt{81 - x^2}}}{81 - x^2} = \frac{\sqrt{81 - x^2} + \frac{x^2}{\sqrt{81 - x^2}}}{81 - x^2} = \frac{81}{(81 - x^2)^{3/2}}$$.Evaluating at $x = 0$, we get $f'(0) = \frac{81}{81^{3/2}} = \frac{1}{9}$.3. The second derivative $f''(x)$ is more complex, but we don't need it for the linear (first-order) term in the quadratic approximation.Thus, the quadratic approximation at $x = 0$ is $Q(x) = f(0) + f'(0)x = 0 + \frac{1}{9}x = \frac{x}{9}$.

Find the Quadratic Approximation of F at X = 0 f(x)=x81−x2f ( x ) = \frac { x } { \sqrt { 81 - x ^ { 2 } } }f(x)=81−x2​x​

Find the Quadratic Approximation of F at X = 0 $f ( x ) = \frac { x } { \sqrt { 81 - x ^ { 2 } } }$