Choosing a Linearization: Tangent Line vs. Secant
A physics lab needs to estimate sqrt(4.1) without a calculator. Two students propose methods. Student A uses the tangent line to f(x)=sqrt(x) at x=4. Student B uses the secant line through (4,2) and (9,3). A third student suggests using the tangent line at x=9 instead, arguing that 4.1 is 'closer to 4' so it does not matter. The lab supervisor asks the group to justify which linear approximation is appropriate for x=4.1, to bound the error qualitatively, and to explain what property of f makes the tangent-line approximation good near the point of tangency. The supervisor also notes that the derivative of sqrt(x) is unbounded near 0 and asks whether that changes the reasoning.
1. Compute both approximations for sqrt(4.1) and compare them to a sensible reference. Which is better and why?
Read the worked answer
Tangent at x=4: f(4)=2, f'(x)=1/(2 sqrt x), so f'(4)=1/4. Linearization L(x)=2+(1/4)(x-4). At x=4.1: L(4.1)=2+(1/4)(0.1)=2+0.025=2.025.
Secant through (4,2),(9,3): slope = (3-2)/(9-4)=1/5=0.2. Line: S(x)=2+0.2(x-4). At 4.1: S(4.1)=2+0.2(0.1)=2.02.
Reference: sqrt(4.1) is slightly above 2.0248. The tangent value 2.025 is very close; the secant gives 2.02, farther off. The tangent at x=4 is the better approximation because 4.1 is near 4, where the tangent matches both value and slope.
2. Why is the tangent line at x=9 a poor choice for estimating sqrt(4.1), even though 4.1 is 'near' 4? Frame your answer in terms of differentiability and local behavior.
Read the worked answer
The tangent at x=9 is L9(x)=3+(1/6)(x-9). At x=4.1: L9(4.1)=3+(1/6)(-4.9)=3-0.8167=2.1833, which is far above the true 2.0248. The tangent line is a local approximation: its accuracy degrades with distance from the point of tangency, and here we are extrapolating about 4.9 units away. The error is roughly (1/2) f''(c)(x-a)^2, and with |x-a| large the quadratic term is large. Also f''(x) = -1/(4 x^{3/2}) is small in magnitude near 9 but the distance dominates. The relevant criterion is proximity to the expansion point, not the absolute location of 4.1.
3. The supervisor raises the unbounded derivative of sqrt(x) near 0. Explain why the tangent-line method still works at x=4 but would be problematic near x=0.
Read the worked answer
f'(x)=1/(2 sqrt x) blows up as x->0+, so f is not differentiable at 0 (vertical tangent) and the derivative is huge near 0. At x=4 the derivative is finite (1/4) and f'' is finite, so the linearization has a controlled error term. Near 0, the slope changes extremely rapidly, so a tangent line at, say, x=0.01 would have a very steep slope and would poorly approximate nearby values; the curvature is large. The method's validity rests on f being differentiable with bounded curvature near the base point, which holds at 4 but fails near 0.