# Use linear approximation, i.e. the tangent line, to approximate 125.1^(1/3) as follows: Let f(x)=x^(1/3).The equation of the tangent line to f(x) at x=125 can be written in the form y=mx+b where m...

Use linear approximation, i.e. the tangent line, to approximate 125.1^(1/3) as follows: Let f(x)=x^(1/3).

The equation of the tangent line to f(x) at x=125 can be written in the form y=mx+b where

m =

b =

Using this, we find our approximation for 125.1^(1/3) is ______

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Use a linear function to approximate `125.1^(1/3)` :

(1) Let `f(x)=x^(1/3)`

(2) We will find a linear approximation to f(x) at x=125.

(3) `f(125)=5`

(4) `f'(x)=(1/3)x^(-(2)/3)` . This is the slope of the tangent line to f(x); at x=125 we have `f'(125)=1/3*125^(-2/3)=1/3*1/25=1/75`

(5) The equation of the line tangent to f(x) at x=125 is the linear approximation to f(x) at x=125. The equationof the line is:

`y-5=1/75(x-125)` or `y=1/75x+10/3`

**So m=1/75 and b=10/3**

`f(125.1)~~1/75(125.1)+10/3 = 5.001bar(3)`

** actual value: `125.1^(1/3)~~5.00133297794`