Sign Up For Free

**Join for FREE**,
and start talking with other members, weighing in on community polls,
and more.

Also, by registering and logging in you'll see fewer ads and pesky welcome messages (like this one!)

- Reply to threads, and start your own
- Create reports of your
**campus visits** - Share college
**photos**and**videos** **Find your dream college**, save your search and share with friends- Receive our
**monthly newsletter**

Home
/
College Discussion / College Admissions and Search / SAT and ACT Tests & Test Preparation / AP Tests Preparation

College Confidential’s “Dean,” Sally Rubenstone, put together 25 of her best tips. So far, the "25 Tips from the Dean" eBook has helped more than 10K students choose a college, get in, and pay for it. Get your free copy: http://goo.gl/9zDJTM

SATGamer
Registered User Posts: **11** New Member

I don't understand this question.

The approximate value of y = √(4+sin x) at x = 0.12, obtained from the tangent to the graph at x = 0 is

A) 1.98

B) 2

C) 2.03

D) 2.24

E) 3

I just plugged 0.12 into the equation and got 2 which is B, but that is the wrong answer according to my answer key.

How do I get the right answer?

The approximate value of y = √(4+sin x) at x = 0.12, obtained from the tangent to the graph at x = 0 is

A) 1.98

B) 2

C) 2.03

D) 2.24

E) 3

I just plugged 0.12 into the equation and got 2 which is B, but that is the wrong answer according to my answer key.

How do I get the right answer?

Post edited by SATGamer on

## Replies to: Help: How do I solve this AP Calculus question?

2,118Senior Membery(0.12) ≈ y(0) + (0.12)y'(0)

= 2 + (0.12)y'(0)

By chain rule, y'(x) is equal to (1/2)(4 + sin x)^(-1/2) (cos x), in which y'(0) = 1/4 or 0.25. Then

y(0.12) ≈ 2 + (0.12)(0.25) = 2.03, C.

938Member260Junior Membery(point being approximated) = y(tangent point) + (point being approximated)y'(tangent point)

C is the correct answer.

Btw, is there suppose to be a rule that shows that the value is a little more than 2?

2,118Senior MemberFor small Δx, the following expression gives usually a good approximation:

y(x0 + Δx) ≈ y(x0) + Δx*y'(x0), given that y is differentiable. This formula generalizes to higher dimensions.

y(0.12) is greater than 2 because sin(0.12) > 0, so sqrt(4 + sin(0.12)) > sqrt(4).

11New Member12New MemberThe only "rule" I know about these approximations is that if the function is concave down in the area you're looking at, the linear approximation will be an overestimate, and if the function is concave up, the linear approximation will be an underestimate.