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SATGamer
Posts: **11**Registered User 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

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

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?

2,118Registered User Senior 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.

938Registered User Member260Registered User Junior 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?

2,118Registered User Senior 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).

11Registered User New Member12Registered User New 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.