Linear approximation and differentials: Question bank

Link back to the parent: Linear approximation.

True or false — justify

L (x)

and

f (x)

always agree exactly at

x = a

True.

L (a) = f (a) + f^{'} (a) (a - a) = f (a)

, the

(x - a)

run is zero, so the line and curve touch. The approximation is exact only at the base point.

The approximation

f (x) \approx L (x)

gets worse the further

x

is from

a

True in general. The leftover error is

ε (x) (x - a)

; as

x

moves away,

(x - a)

grows and

ε (x)

stops being tiny, so the straight line drifts off the curving graph.

f

is itself a straight line, then

L (x) = f (x)

for all

x

, not just near

a

True. A line's tangent is the line itself; the error term

ε (x) (x - a)

is exactly

0

everywhere because a line has no curvature to peel away.

d y

and

Δ y

are the same number.

False.

d y

is the rise along the tangent;

Δ y

is the true rise of the curve. They coincide only for a straight

f

; otherwise their gap is the curvature error.

You can make the linearization error zero by choosing a smart base point

a

False (for a genuinely curved

f

). A better

a

shrinks the error, but as long as the curve bends between

a

and

x

, a straight line cannot match it exactly.

f^{'} (a) = 0

, linear approximation is useless.

False. It gives

L (x) = f (a)

, a flat estimate — perfectly valid, just says "

f

is roughly constant here." It's only crude, because the leading behaviour near a flat point is quadratic, handled by Concavity and second derivative.

Doubling the step

d x

roughly doubles the approximation error.

False. The error behaves like

(x - a)^{2}

(the next Taylor term), so doubling the step roughly quadruples the error. See Taylor series.

d x

must be positive.

False.

d x

is any chosen change in

x

; it can be negative (target to the left of

a

) or zero. A negative

d x

just gives a negative run and a downhill/uphill rise accordingly.

The formula

Δ y \approx d y

holds only when

f

is increasing.

False. It holds near any point where

f

is differentiable, increasing or decreasing. The sign of

f^{'} (a)

only sets whether the tangent rises or falls.

Spot the error

4.1 \approx 2 + \frac{1}{4} = 2.25

Error: dropped the run. The rise is slope × run

= \frac{1}{4} \cdot (4.1 - 4) = \frac{1}{4} (0.1) = 0.025

, giving

2.025

. Multiplying the slope by

1

instead of

0.1

inflates the answer.

f (x) \approx f (a) + f^{'} (a)

, so I just add the slope to the value."

Error: units/dimensions mismatch. You must multiply slope by the run

(x - a)

. Slope has units of rise-per-run; alone it isn't a rise. The correct form is

f (a) + f^{'} (a) (x - a)

"To estimate

4.1

I'll linearize

x

a = 0

because

0 = 0

is easy."

Error: bad base point.

f^{'} (0) = \frac{1}{2 0}

is undefined (infinite slope), and

4.1

is far from

0

. Choose the nearest easy exact point,

a = 4

d V = 4 π r^{2} d r

, and since

d r = 0.05

the true volume error is exactly

20 π

Error: 'exactly'.

d V

is the tangent-line estimate of the error, so

Δ V \approx d V

, not

=

. The genuine change

\frac{4}{3} π [(r + d r)^{3} - r^{3}]

carries extra

r d r^{2}

and

d r^{3}

terms.

(1 + x)^{n} \approx 1 + n x

works for any

x

Error: only for small $x$ . It's the linearization at

a = 0

; away from

0

the neglected

(2 n) x^{2}

term matters. Try

x = 1, n = 10

: formula gives

11

, truth is

1024

"Since

sin x \approx x

, I can use it to estimate

sin (3)

Error: $x$ isn't small. The approximation is the tangent at

0

;

x = 3

radians is far away where

sin

has curved right past.

sin 3 \approx 0.141

, not

3

e^{0.5} \approx 1 + 0.5 = 1.5

is a great estimate."

Error: overclaimed accuracy. It's the tangent-at-0 value;

e^{0.5} = 1.6487 \dots

, so it's off by about

0.15

— usable but not "great," because

0.5

isn't tiny. The quadratic term adds

\frac{0.25}{2} = 0.125

Why questions

Why do we split the difference quotient as

f^{'} (a) + ε (x)

rather than just say it "equals"

f^{'} (a)

Because for

x \neq = a

the quotient is only close to

f^{'} (a)

, not equal; the term

ε (x)

names that gap, and it

\to 0

x \to a

by the limit definition.

Why does the error shrink faster than

(x - a)

itself?

The error is

ε (x) \cdot (x - a)

— a small factor

ε (x) \to 0

times a small run — so it's a product of two shrinking things, vanishing faster than either alone.

Why choose the nearest base point where

f (a)

is exact?

Nearest keeps the run

(x - a)

tiny, which shrinks the

(x - a)^{2}

-sized error; exactness keeps

f (a)

and

f^{'} (a)

from contributing their own errors.

Why is the tangent line the best straight-line approximation, not just a good one?

It's the unique line matching both

f (a)

(value) and

f^{'} (a)

(slope); any other line disagrees in slope, giving error growing linearly in

(x - a)

instead of quadratically.

Why write

d y = f^{'} (x) d x

instead of just using

Δ y

directly in error propagation?

Because in a lab you don't know the true error, only a small bound

d x

on the input;

d y

converts that known input tolerance into an output tolerance via the slope. See Error propagation.

Why does linearizing

(1 + x)^{n}

a = 0

give such a clean formula?

Setting

a = 0

collapses the run

(x - a)

to just

x

, and

f (0) = 1, f^{'} (0) = n

, so

L (x) = 1 + n x

with nothing to carry.

Why can Newton's method be seen as repeated linear approximation?

Each step replaces

f

by its tangent line and solves the line

= 0

; the root of the line is the next guess, so Newton's method is linear approximation iterated toward a zero.

Edge cases

What does linear approximation give when

f

has a vertical tangent at

a

(slope infinite)?

It fails —

f^{'} (a)

doesn't exist, so

L (x)

is undefined. Example:

x

a = 0

. Pick a base point where the slope is finite.

What happens if

f

is not differentiable at

a

(e.g. a corner like

∣ x ∣

0

No single tangent line exists, so there's no linearization at that point. Left and right slopes disagree; you'd need to pick a base point on a smooth stretch.

If the target equals the base point (

x = a

, so

d x = 0

), what does the estimate give?

Exactly

f (a)

, with zero error — the run is zero, so line and curve coincide. A degenerate but perfectly correct case.

For a concave-up function, is

L (x)

an over- or under-estimate near

a

Under-estimate: a concave-up curve bends above its tangent, so the line sits below the true value. Concave-down flips this — the line over-estimates. Governed by Concavity and second derivative.

ε (x)

doesn't go to zero as

x \to a

, what has gone wrong?

Then

f^{'} (a)

wasn't actually the derivative — the difference quotient isn't converging to it.

ε (x) \to 0

is precisely the statement that the limit defining

f^{'} (a)

exists.

What if you take

d x

enormous — is

d y

still meaningful?

d y = f^{'} (a) d x

is always defined (rise on the tangent), but it stops approximating

Δ y

: for large

d x

the curve has bent far from the line, so

Δ y \neq \approx d y

Connections

Derivative as a limit — the $ε (x) \to 0$ statement these traps hinge on.
Tangent line — the geometric object every "line vs curve" question is about.
Concavity and second derivative — decides over- vs under-estimate.
Taylor series — the $(x - a)^{2}$ error term named in several answers.
Newton's method — repeated linearization.
Error propagation — why $d y = f^{'} (x) d x$ beats using $f$ directly.