calculus
posted by jin .
se differential, i.e., linear approximation, to approximate (8.4)^(1/3) as follows:
Let f(x)=(x )^(1/3). The linear approximation to f(x) at x=8 can be written in the form y=mx+b where m is: and where b is:
Using this, we find our approximation for (8.4)^(1/3) is

calculus 
Damon
y = x^(1/3)
dy/dx = (1/3) x^(1/3)
at x = 8
y(8) = 2
dy/dx = (1/3)/2
y(x+h) = y(x) + h dy/dx
y(8.4) = 2 + .4 (1/3)/2
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