Asked by Chris
Use the fact that (125)^1/3 = 5 and the tangent line approximation to estimate (123)^1/3
Answers
Answered by
Reiny
Let the function be y = x^(1/3)
so we know (125, 5) lies on it, we need (123, ?)
dy/dx = (1/3)x^(-2/3)
at (125,5) , dy/dx = 1/75
the tangent equation is y = (1/75)x + b
but (125,5) is on it
5 = (1/75)(125) + b
5 = 5/3 + b
b = 10/3
tangent line at (125,5) is y = (1/75)x + 10/3
so when x = 123
y = 123/75 + 10/3 = 4.9733
(calculator answer for 123^(1/3) is 4.97319
so we know (125, 5) lies on it, we need (123, ?)
dy/dx = (1/3)x^(-2/3)
at (125,5) , dy/dx = 1/75
the tangent equation is y = (1/75)x + b
but (125,5) is on it
5 = (1/75)(125) + b
5 = 5/3 + b
b = 10/3
tangent line at (125,5) is y = (1/75)x + 10/3
so when x = 123
y = 123/75 + 10/3 = 4.9733
(calculator answer for 123^(1/3) is 4.97319
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