let f(x) = x^(1/3) The equation of the tangent line to f(x) at (x) = 125?

Question

can be written in the form.

y= mx+b

where m = ?

b = ?

Using this, we find our approximation for 125.4^(1/3)

Answer 1

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f '(x) = (1/3)x^(-2/3)

f(125) = 5

f '(125) = (1/3)*(1/25)

............ = 1/75

Thus the equation is

y - 5 = (1/75)(x - 125)

y = x/75 + 10/3

The assumption now is that at x = 125.4, the curve will still be very close to the tangent, so we take as an approximation

(125.4)^(1/3) = 125.4/75 + 10/3

..................... = 1.672 + 3.333

...................... = 5.005