Linearization...

**omibayne** · November 30th 2008, 07:40 PM

Find the linearization L(x) of f(x) = tanx at x= pi

**Chris L T521** · November 30th 2008, 07:46 PM

Originally Posted by omibayne

Find the linearization L(x) of f(x) = tanx at x= pi

Recall that $L(x)= f(x_0)+f'(x_0)(x-x_0)$

$\because f(x)=\tan(x),~f'(x)=\sec^2(x)$

Thus, $f(\pi)=\tan(\pi)=0$ and $f'(\pi)=sec^2(\pi)=1$

Therefore, $L(x)=\dots$

Can you take it from here?

**omibayne** · November 30th 2008, 07:57 PM

Originally Posted by Chris L T521

Recall that $L(x)= f(x_0)+f'(x_0)(x-x_0)$

$\because f(x)=\tan(x),~f'(x)=\sec^2(x)$

Thus, $f(\pi)=\tan(\pi)=0$ and $f'(\pi)=sec^2(\pi)=1$

Therefore, $L(x)=\dots$

Can you take it from here?

so is it $x-\pi$ ???

**Chris L T521** · November 30th 2008, 07:58 PM

Originally Posted by omibayne

so is it $x-\pi$ ???

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