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

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May 22, 2016

Note that #f(0)=(1+0)^k=1#.

Assuming #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#.

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