Wait I'm a little confused. So the rule for when you approximate a value with a tangent point is that you're suppose to plug in the point that's tangent (in this case it is 0) instead of the point you are trying to find (in this case it is .12) into the equation and then add that value to the derivative of the equation when plugging in the tangent point times the value you are trying to approximate?
y(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?
@StraferKev, we're trying to approximate y(0.12) by plugging in y(0) and y'(0). These values are easy to find w/out a calculator, while y(0.12) or y'(0.12) are not.
For 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).
@StraferKev, there is no "rule" that says that the value will be a little more than 2. rspence simply plugged 0 into the original equation for x, found it to be 2, and then logically deduced that .12 would produce a slightly larger value than 0 in this situation.
The 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.
Replies to: Help: How do I solve this AP Calculus question?
y(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.
y(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?
For 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).
The 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.