the way I recall linear approximation is
f(x) = f(x0) + f'(x0)(x - x0)
here x = 8.4, x0 = 8
and f'(x) = 1/(3x^(-2/3))
so f(8.4) =(appr.) f(8) + f'(8)(8.4-8)
=appr 2 + (1/12)(.4) = 2 + 1/30
=appr 2.03333
(by calculator 8.4^(1/3) = 2.033
Use linear approximation, i.e. the tangent line, to approximate 8.4^(1/3) as follows:
Let f(x)= x^(1/3) . The equation of the tangent line to f(x) at x=8 can be written in the form y=mx+c where m=1/12 and c=4/3:
Using this, find our approximation for 8.4^(1/3).
2 answers
let's try that again:
the way I recall linear approximation is
f(x) = f(x0) + f'(x0)(x - x0)
here x = 8.4, x0 = 8
and f'(x) = 1/(3x^(-2/3))
so f(8.4) =(appr.) f(8) + f'(8)(8.4-8)
=appr 2 + (1/12)(.4) = 2 + 1/30
=appr 2.03333
(by calculator 8.4^(1/3) = 2.033
the way I recall linear approximation is
f(x) = f(x0) + f'(x0)(x - x0)
here x = 8.4, x0 = 8
and f'(x) = 1/(3x^(-2/3))
so f(8.4) =(appr.) f(8) + f'(8)(8.4-8)
=appr 2 + (1/12)(.4) = 2 + 1/30
=appr 2.03333
(by calculator 8.4^(1/3) = 2.033