What values of x guarantee that f(x) = x^3 is within 0.05 unit of 8?

Select one:

a. If x is within 0.01 unit distance of 2, then f(x) is within 0.05 unit of 8.

b. If x is within 2 units distance of 2, then f(x) is within 0.05 unit of 8.

c. If x is within 1.99 units distance of 2, then f(x) is within 0.05 unit of 8.

d. If x is within 2.01 units distance of 2, then f(x) is within 0.05 unit of 8.

Let h be the difference between x and 2, then f(2+h)=(2+h)³=(2(1+h/2))³=8(1+h/2)³.

This gives us 8(1+3h/2) approximately=8+12h.

So if 12h=0.05 then h=0.05/12=0.004 approx, which means that x must be within 0.004 units of 2.

To prove this we can cube 2.004 and 1.996 and we get roughly 8.048 which is about 8.05, and about 7.952. These are both within 0.05 units of 8, proving that x must be within 0.004 units of 2. The closest answer (not very close!) in the options is a.

Note that if h=0.01, 12h=0.12 so we would actually be within 0.12 units of 8. To prove this, 1.99³=7.88 approx and 2.01³=8.12 approx. Therefore, if x is within 0.01 units of 2 it does NOT guarantee that f(x) will be within 0.05 units of 8; it can only guarantee being within 0.12 units of 8.

by Top Rated User (716k points)