We need to solve √(1+x)=2±0.0002.

There are two ways to do this. First, square both sides:

1+x=(2±0.0002)²=4±0.0008+0.00000004=4.00000004±0.0008.

From this x=3.00000004±0.0008≈3±0.0008. So x is within 0.0008 units of 3.

The other way is to write √(1+x) as √(1+(3+h))=√(4+h)=√(4(1+h/4)).

This is 2√(1+h/4)=2(1+h/4)^½≈2(1+h/8).

So 2(1+h/8)=2±0.0002. Divide through by 2: 1+h/8=1±0.0001 making h=±0.0008, because the 1s cancel out.

If x is within 0.0008 units of 3, then f(x) is within 0.0002 units of 2.

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To help you further with this type of question, let’s use some algebra.

If L is the limit value for x (lim as x➝L=V) and h is the small variation in x and k the small variation in f(x) near the limit, then we can write f(L+h)=V+k. We need to know the definition of f(x) so that we can find the relationship between h and k. We can also write V=f(L) so f(L+h)=f(L)+k. Given h, we can find k, or given k, we can find h, provided we know what f(x) is.

EXAMPLES

(1) f(x)=2x+1, L=3, V=7=f(3)=f(L). f(3+h)=7+k using f(L+h)=f(L)+k.

In algebraic terms, f(L+h)=2(L+h)+1 and f(L)=2L+1. So 2(L+h)+1=2L+1+k. And 2L+1+2h=2L+1+k. Remove 2L+1 from each side: 2h=k or h=k/2. If k=0.54 then h=0.27. If h=0.01 then k=0.02. We see below the same result by substituting numbers.

Substituting for L we have f(3+h)=2(3+h)+1=7+2h; f(3)=2×3+1=7, so 7+2h=7+k and 2h=k or h=k/2. If k is given as 0.54 then h=0.27. Or if h is given as 0.01, then k=2h=0.02.

(2) f(x)=√(x+1), L=3, V=2=f(3)=f(L).

Stick with algebra for the moment: f(L+h)=√(L+h+1). Also f(L)=√(L+1).

f(L+h)=f(L)+k, so √(L+h+1)=√(L+1)+k.

We can write the left-hand side √((L+1)(1+h/(L+1)).

This becomes √(L+1)(1+h/(L+1))^½=√(L+1)(1+h/(2(L+1)) approximately when h is small. This expands to √(L+1)+h/(2√(L+1)).

So, √(L+1)+h/(2√(L+1))=√(L+1)+k. We can remove √(L+1) from each side:

h/(2√(L+1))=k which relates h and k. So we also write h=2k√(L+1). Now we can start to substitute values for L and h or k.

If we are given k=0.0002 then h=0.0004√(3+1)=0.0004×2=0.0008.