How do you show that the linearization of $f (x) = {(1 + x)}^{k}$ $f(x) = (1+x)^k$ at x=0 is $L (x) = 1 + k x$ $L(x) = 1+kx$ ?

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Answer 1 · 2016-05-22T16:04:00

Note that $f (0) = {(1 + 0)}^{k} = 1$ $f(0)=(1+0)^k=1$ .

Assuming $k$ $k$ is a constant, we see that $f'(x)=k(1+x)^(k-1)$ . Thus $f'(0)=k(1+0)^(k-1)=k*1=k$ .

Using the point $(0,1)$ and slope of $k$ we can write the linearization function: $L(x)=k(x+0)+1=1+kx$ .

How do you show that the linearization of f(x) = (1+x)^kf(x)=(1+x)kf(x) = (1+x)^k at x=0 is L(x) = 1+kxL(x)=1+kxL(x) = 1+kx?