f(x) = ln(1 + x); a = 0; f(0.9)

Question

Please help ASAP sirs thanks alot

Answer 1

1. Differentiate f(x); so f'(x)=1/(1+x). f'(0)=1 which is the gradient of the line of the tangent.

y=L(x)=x+b where b is found by substituting x and y values. We need f(0)=ln(1)=0, so the (x,y) values will be (0,0). That makes b=0. So y=L(x)=x. Nice and simple!

2. 0.9 is close to 0 so L(0.9)=0.9.

3. f(0.9)=ln(1.9)=0.6419, so the approximation is quite poor. The %ge error is 100(0.9-ln(1.9))/ln(1.9)=40.22% too high.