<p>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</p>
<p>A) 1.98
B) 2
C) 2.03
D) 2.24
E) 3</p>
<p>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?</p>
<p>Of course the actual value of y is going to be a little more than 2. To approximate y(0.12), we find y(0) first and use</p>
<p>y(0.12) ≈ y(0) + (0.12)y’(0)</p>
<p>= 2 + (0.12)y’(0)</p>
<p>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</p>
<p>y(0.12) ≈ 2 + (0.12)(0.25) = 2.03, C.</p>
<p>Edit: Nevermind, I was beat to it lol.</p>
<p>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? </p>
<p>y(point being approximated) = y(tangent point) + (point being approximated)y’(tangent point)</p>
<p>C is the correct answer.</p>
<p>Btw, is there suppose to be a rule that shows that the value is a little more than 2?</p>
<p>@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.</p>
<p>For small Δx, the following expression gives usually a good approximation:</p>
<p>y(x0 + Δx) ≈ y(x0) + Δx*y’(x0), given that y is differentiable. This formula generalizes to higher dimensions.</p>
<p>y(0.12) is greater than 2 because sin(0.12) > 0, so sqrt(4 + sin(0.12)) > sqrt(4).</p>
<p>Ahh I get it now, I think I was thrown off by the wording. Thank you rspence, it makes a lot of sense now.</p>
<p>@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.</p>