Linear approximation, question 12: UBC MATH 100 practice

Question 12 of 15Challenging3 marks · 6 minTheory UBC Math 100 §9.2CLP-1 §3.4Key results

A drug’s effect at concentration $c$ (in mg/L), as a fraction of its greatest possible effect, is modelled by $E(c) = \dfrac{c^{2}}{1 + c^{2}}$ . The tangent line at $c = 1$ gives the estimate $E(1.2) \approx 0.6$ . Which statement is correct?

It is an overestimate, because E″(1) < 0: near c = 1 the graph bends down, below its tangent line.
It is an underestimate, because E″(1) < 0, and a concave-down graph lies above its tangent line.
It is an underestimate, because E′(1) > 0: the effect is rising, so the graph climbs above its tangent line.
It is an underestimate, because E″(1) > 0: near c = 1 the graph bends up, above its tangent line.
It is an overestimate, because E″(1) > 0: near c = 1 the graph bends up, and a concave-up graph lies below its tangent line.

How sure are you?

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Theory UBC Math 100 §9.2 CLP-1 §3.4Key results

Worked solution

Strategy

Whether the graph lies above or below its tangent line is a question about bending, so it is settled by the sign of the second derivative on the step from 1 to 1.2, not by the first derivative.

Working

$E'(c) = \frac{2c(1 + c^{2}) - c^{2}\cdot 2c}{(1 + c^{2})^{2}} = \frac{2c}{(1 + c^{2})^{2}}$
Quotient rule; the c³ terms cancel.
$E''(c) = \frac{2(1 + c^{2})^{2} - 2c\cdot 2(1 + c^{2})\cdot 2c}{(1 + c^{2})^{4}} = \frac{2(1 - 3c^{2})}{(1 + c^{2})^{3}}$
Quotient rule again, then cancel one factor of 1 + c².
$E''(c) < 0 \text{ for } c > \tfrac{1}{\sqrt{3}}, \qquad E''(1) = \frac{2(-2)}{8} = -\frac{1}{2}$
So E is concave down on the whole step from 1 to 1.2, not just at 1.
$E(1.2) = \frac{1.44}{2.44} \approx 0.590 < 0.6$
A concave-down graph lies below its tangent lines, so the estimate is too high, as the true value confirms.

Answer

\text{An overestimate: } E''(1) = -\frac{1}{2} < 0

Check

At c = 1.1 too, E = 1.21/2.21 ≈ 0.548 is below the tangent line’s 0.55, so the curve sits under its tangent there.

Where this goes wrong

Reading the picture the wrong way round. A concave-down graph bends away below its tangent line, so a tangent-line estimate is too high; concave up, too low.