Question

Linear Approximation (.99)^25

Answer 1

think of f(x)=x^25
f'(x)=25x^24
linear approximation
f(25)= f(1)+ f'(1)(0.99-1)=1-25*0.01=0.75
which is fairly close to the actual value: 0.777821

Answer 2

f(25) that is f(1)

Answer 3

Andras has it. you could also use \[f(x)=(1-x)^{25}\]
\[f'(x)=-25(1-x)^{24}\] find the equation of the line tangent to the curve and then let x = .1

Answer 4

Thanks!!! You guys are awesome!! :D