Find the linearization

Question

Find the linearization of the function f(x)=x6 at x=1

L(x) =

Using this linearlization, f(1.05).is approximately =

Answer 1

f(x)=x⁶

f'(x)=6x⁵

L(x)=f(a)+f'(a)(x-a)

a =1

L(x)=f(1)+f'(1)(x-1)=1+6(x-1)

L(1.05)=1+6(1.05-1)=1+6(.05)=1+0.3=1.3

Note: f(1.05) = 1.05⁶ = 1.340095641