Find the quadratic approximation of f at x = 0.
-f(x) = tan 10x
A) $Q ( x ) = 1 + 10 x ^ { 2 }$
B) $Q ( x ) = 1 - 5 x ^ { 2 }$
C) $\mathrm { Q } ( \mathrm { x } ) = 1 + 5 \mathrm { x } ^ { 2 }$
D) $Q ( x ) = 10 x$

Question

Find the quadratic approximation of f at x = 0.
-f(x) = tan 10x 
A) $Q ( x ) = 1 + 10 x ^ { 2 }$
B) $Q ( x ) = 1 - 5 x ^ { 2 }$
C) $\mathrm { Q } ( \mathrm { x } ) = 1 + 5 \mathrm { x } ^ { 2 }$
D) $Q ( x ) = 10 x$

Quizplus · Accepted Answer

The quadratic approximation of a function $f$ at $x = a$ is given by $f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$. For $f(x) = \tan(10x)$ at $x = 0$, the first derivative $f'(x) = 10\sec^2(10x)$ and $f'(0) = 10$, and the second derivative $f''(x)$ involves $\sec^2(10x)$ and $\tan(10x)$ terms, but since we're evaluating at $x = 0$, the quadratic term vanishes because $f''(0)$ is multiplied by $(x - 0)^2/2$, and $\tan(0) = 0$, making the constant and quadratic terms 0. Thus, the approximation is linear, $10x$.

Find the Quadratic Approximation of F at X = 0 Q(x)=1+10x2Q ( x ) = 1 + 10 x ^ { 2 }Q(x)=1+10x2

Find the Quadratic Approximation of F at X = 0 $Q ( x ) = 1 + 10 x ^ { 2 }$