Given #f(x)=sqrtx# when x=25, how do you find the linear approximation for #sqrt25.4#?
1 Answer
Sep 17, 2017
Explanation:
Given:
#f(x) = sqrt(x) = x^(1/2)#
We have:
#f(25) = sqrt(25) = 5#
#f'(x) = 1/2 x^(-1/2)#
#f'(25) = 1/2 1/sqrt(25) = 1/10#
So:
#f(25.4) = f(25+0.4) ~~ f(25) + 0.4*f'(25) = 5+0.4/10 = 5.04#
