Linear Approximation (.99)^25

Question

anonymous · Answer

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.

anonymous · Answer

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.

anonymous · Answer

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