Question
find the linear approximation to f(x)=1/square root of(x^3+1) at a=2
Answers
Steve
f(x) = (x^3+1)^(-1/2)
f'(x) = -3/2 x^2 (x^3+1)^(-3/2)
f(2) = 1/3
f'(2) = -2/9
the linear approximation is just the tangent line at x=2, so
y - 1/3 = -2/9 (x-2)
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D1%2F%E2%88%9A(x%5E3%2B1),+y+%3D+-2%2F9+(x-2)+%2B+1%2F3
f'(x) = -3/2 x^2 (x^3+1)^(-3/2)
f(2) = 1/3
f'(2) = -2/9
the linear approximation is just the tangent line at x=2, so
y - 1/3 = -2/9 (x-2)
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D1%2F%E2%88%9A(x%5E3%2B1),+y+%3D+-2%2F9+(x-2)+%2B+1%2F3