Approximating Values of a Function Using Local Linearity and Linearization

AP Calculus · Contextual Applications of Differentiation

Local Linearity and Linearization

Use the tangent line in place of the curve — and let concavity tell you which way you erred.

key idea$L(x)=f(a)+f'(a)(x-a)\approx f(x)\ \text{ for }x\text{ near }a$

Setting up the model

Zoom far enough into a differentiable curve and it looks straight. Local linearity is that observation; linearization is using the tangent line as a stand-in for the function near the point of tangency.

Building and using the linearization

Choose $a$ — a nearby input where $f(a)$ and $f'(a)$ are easy to compute exactly.
Compute $f(a)$ and $f'(a)$. These are the only two numbers the tangent needs.
Write $L(x)=f(a)+f'(a)(x-a)$ — point-slope form of the tangent line.
Evaluate $L$ at the target, not $f$. That is the whole point: $f$ is hard, $L$ is linear.
Decide over or under from the concavity: concave up ⇒ under-estimate, concave down ⇒ over-estimate.

concave down → tangent OVER-estimates

concave up → tangent UNDER-estimates

The tangent at the marked point. On a concave-down curve the tangent lies above, so it over-estimates; on a concave-up curve it lies below and under-estimates. The sign of $f''$ decides, not the sign of $f'$.

Worked examples

EX 1

Approximate a root

$\sqrt{9.2}$

Use a linearization to approximate $\sqrt{9.2}$.

Choose the anchor

$f(x)=\sqrt{x}$ with $a=9$ — the nearest input with an exact root.

Compute the two numbers

$f(9)=3$ and $f'(x)=\dfrac{1}{2\sqrt x}$, so $f'(9)=\dfrac16$.

Write the linearization

$L(x)=3+\dfrac16(x-9)$.

Evaluate at the target

$L(9.2)=3+\dfrac16(0.2)=3+0.0333\ldots\approx3.0333$.

Compare

The true value is $3.03315\ldots$ — accurate to four decimals from one line of arithmetic.

EX 2

Over or under?

concavity decides

Is the estimate in Example 1 too large or too small? Justify without computing the true value.

Find the second derivative

$f'(x)=\tfrac12 x^{-1/2}$, so $f''(x)=-\tfrac14 x^{-3/2}$.

Check its sign near $a=9$

$f''(9)=-\tfrac{1}{108}\lt 0$, so $f$ is concave down there.

Apply the rule

A concave-down curve lies below its tangent, so the tangent value is too big.

Conclude

$L(9.2)$ is an over-estimate of $\sqrt{9.2}$.

Verify

$3.03333\gt 3.03315$. ✓ The justification is what earns the mark, not the comparison.

EX 3

From a table

no formula

$g$ is differentiable with $g(4)=7$ and $g'(4)=-2$. Estimate $g(4.3)$.

Identify the anchor

$a=4$, the point where both values are known.

Write the linearization

$L(x)=7-2(x-4)$.

Evaluate

$L(4.3)=7-2(0.3)=7-0.6=6.4$.

State the estimate

$g(4.3)\approx6.4$.

Note what is unknown

Without $g''$ you cannot say whether this is an over- or under-estimate. Only concavity settles that, and none was given.

EX 4

How far can you push it?

error growth

Using $L(x)=3+\tfrac16(x-9)$ for $\sqrt{x}$, compare the estimates at $x=9.2$ and $x=16$.

Near the anchor

$L(9.2)\approx3.0333$ against the true $3.0332$ — an error of about $0.0002$.

Far from the anchor

$L(16)=3+\tfrac16(7)\approx4.167$ against the true $4$ — an error of $0.167$.

Compare

Moving $35$ times further from $a$ multiplied the error by roughly $1000$.

Explain

Linearization error grows like the square of the distance from $a$, so it degrades fast.

The practical rule

Choose $a$ as close to the target as you can while keeping $f(a)$ and $f'(a)$ exact.

Units & interpretation

Concavity near $a$	Tangent sits	Estimate is
$f''\gt 0$ — concave up	below the curve	an under-estimate
$f''\lt 0$ — concave down	above the curve	an over-estimate
$f''=0$ at an inflection	crosses the curve	could be either — no conclusion
$f''$ unknown	—	cannot decide; say so

Note that $f'$ plays no part in this table. Whether the function is rising or falling is irrelevant to over/under.

Watch out

Concavity decides over/under, not slope

A rising concave-down curve and a falling concave-down curve both give over-estimates. Look at $f''$, never $f'$.

Evaluate $L$, not $f$

The whole point is that $f$ is hard to evaluate at the target. Substituting into $f$ defeats the exercise.

Anchor at a point you know exactly

$a=9$ works for $\sqrt{9.2}$ because $\sqrt9$ is exact. Choosing $a=9.1$ would require knowing $\sqrt{9.1}$ — circular.

Accuracy collapses with distance

The error grows roughly as $(x-a)^{2}$. A linearization is a local tool; far from $a$ it is worthless.

Course map. Topic 4.6 · LO CHA-3.F — approximate a value on a curve using the tangent line (EK CHA-3.F.1, CHA-3.F.2).