Please help ASAP. Thanks!

asked Jul 28 in Calculus Answers by 1998cm Level 1 User (940 points)

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Best answer

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.

answered Jul 28 by Rod Top Rated User (582,800 points)
selected Jul 28 by 1998cm

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.

If instead we had been given h=0.001, then k=0.001/(2×2)=0.001/4=0.00025.

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
81,657 questions
85,896 answers
2,188 comments
69,324 users