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

Question

21207 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-22T16:04:00

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: $L(x)=k(x+0)+1=1+kx$ .

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