Use linear equation to estimate e^0.06. Choose a value of 'a' to produce a small error.

Question

Note: Answers should be in decimal form. Up to two decimal places only.

Answer 1

Let f(x)=eˣ, then f(0)=1. Let a=0.

f'(x)=eˣ and when x=a=0, f'(0)=1 so the gradient at x=0 is 1. This is the gradient of the linear approximation L(x)-f(0)=(1)(x-0), L(x)=1+x. L(0.06)=1.06 which approximates to e^0.06.