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Question

Find the linearization L(x) of ƒ(x) at x = a. ƒ(x) = tan x, a = π

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To find the required linearization, we need to find f(π)f'(\pi), where f(x)=tanxf(x)=\tan x

f(x)=d(tanx)dxf'(x)=\dfrac{d\left(\tan x\right)}{dx}

Remember that: the derivative of tanx\tan x is sec2x\sec^2x

f(x)=sec2xf'(x)=\sec^2x

Substitute x=πx=\pi

f(π)=sec2πf'(\pi)=\sec^2\pi

f(π)=(1)2=1f'(\pi)=(-1)^2=1

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