SOLUTION: Consider the function below. lim x -> 3 x^2=9 What values of x guarantee that f(x) = x^2 is within 0.2 units of 9? If x is within _____ units of 3, then f(x) is within 0.2 unit

Algebra ->  Test -> SOLUTION: Consider the function below. lim x -> 3 x^2=9 What values of x guarantee that f(x) = x^2 is within 0.2 units of 9? If x is within _____ units of 3, then f(x) is within 0.2 unit      Log On


   

Question 1120562: Consider the function below.
lim x -> 3 x^2=9
What values of x guarantee that f(x) = x^2 is within 0.2 units of 9?
If x is within _____ units of 3, then f(x) is within 0.2 units of 9.
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!





John

My calculator said it, I believe it, that settles it

Answer by ikleyn(52779) About Me  (Show Source):
You can put this solution on YOUR website!
.
You want to have, as the problem requires


    |x^2 -9|  <=  0.2.


It means


    -0.2 <= x^2 - 9 <= 0.2


    9-0.2 <= x^2 <= 9+0.2


    8.8   <= x^2 <= 9.2.      Now take the square root from both sides


    sqrt%288.8%29 <= x <= sqrt%289.2%29       (1)


Notice that  sqrt%288.8%29 = 2.966. . .   and  sqrt%289.2%29 = 3.033.


Therefore,  inequality (1)  implies


     2.966 <= x <= 3.033    (with 3 decimals after the decimal point).


It means that  |x - 3|  <= 0.033.


Answer.  If x is within  0.033  units of 3, then f(x) is within 0.2 units of 9.


Notice that sqrt%280.2%29 = 0.447 (approximately.

Therefore, it is NOT ENOUGH to have |x-3| <= sqrt(0.2) in order for the required inequality was in place.

It shows that the solution by the other tutor is incorrect.