Question

Linear Approximation (.99)^25

Answer 1

We know that \[1^{25}=1\]We use the linear approximation as follows:\[f(x) \approx f(a)+f'(a)(x-a)\]where a=1. \[f(0.99)\approx f(1)+f'(1)(-0.01)\]\[f(0.99)\approx 1-0.01(25)\]\[f(0.99) \approx 0.75\]Actual value (to four decimals) is 0.7782. Not bad.

Answer 2

In case you're unsure where I got the terms from. The general function we are using to model this situation is:\[f(x)=x^n\]n, real. We use a known value of this function f(1), in this case, to estimate the value of an unknown value of this function f(0.99). So x=0.99, a=1 and f'(1) is the value of the first derivative of f(x)=x^n, at the point x=1.

Answer 3

Sorry I meant \[f(x)=x^{25}\] is our function. My bad.