apparently you mean
f(x) = ∛x
f(125) = 5
f'(x) = 1/3∛x^2
f'(125) = 1/75
So, now you have a point and a slope. The line is
y-5 = 1/75 (x-125)
Let f(x) = \sqrt[3] x. The equation of the tangent line to f(x) at x = 125 is
y =
Using this, we find our approximation for \sqrt[3] {125.4} is =
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