# Verify the given linear approximation at a=0. Then determine the values of x for which the linear approximation is accurate to within 0.1 (i) 3root(1-x) = 1-(1/3)x (ii) tan x = x (iii) 1/(1+2x)^4 =...

Verify the given linear approximation at a=0. Then determine the values of x for which the linear approximation is accurate to within 0.1

(i) 3root(1-x) = 1-(1/3)x

(ii) tan x = x

(iii) 1/(1+2x)^4 = 1-8x

(iv) e^x = 1 + x

I understand verifying the linear approximation, I am just confused on how to determine the values for x thats accurate within 0.1. Any help would be greatly appreciated!!!!

### 1 Answer | Add Yours

Once you have verified the linear approximation, to find the values for x that make the linear approximation within .1 of the actual value:

Let f(x) be the sactual value and A(x) be the approximation. Then we want `|f(x)-A(x)|<=.1` or `-.1<f(x)-A(x)<.1`

You can solve either with a graphing utility. For example find the values of x so that `1-1/3x` is within .1 of the actual value of `root(3)(1-x)` :

You can graph both the absolute value and the line y=.1 and find the intersections:

The approximate values are x=-1.204336 and x=.706649

Alternatively you can work algebraically:

`-.1<=(1-x/3)-root(3)(x)<=.1`

Working on the right side first:

`1-x/3-root(3)(1-x)<=.1`

`-x/3-root(3)(1-x)<=-.9`

`x+3root(3)(1-x)>=2.7`

`3root(3)(1-x)>=2.7-x`

`27(1-x)>=19.683-21.87x+8.1x^2-x^3`

`x^3-8.1x^2-5.13x+7.317=0`

Again using a algebra utility we find `x~~-1.204336,.706649`

You would use a similar procedure for the rest of the problems:

Find x so that |tanx-x|<.1 etc...