Single-Variable-Calculus · Unit 7 · Video 1 · Interactive Practice

Linear Approximation: The Tangent Line as an Approximation

IKey Formulas

Formula	Name	What it says
$f(x) \approx f(x_0) + f'(x_0)(x - x_0)$	Linear approximation	Near the base point $x_0$ , the curve is approximately its tangent line
$\dfrac{\Delta f}{\Delta x} \approx f'(x_0)$	Difference-quotient form	For small $\Delta x$ , average rate (secant slope) $\approx$ infinitesimal rate (tangent slope)
$\Delta f \approx f'(x_0)\,\Delta x$	Change form	Multiply by $\Delta x$ ; with $\Delta x = x - x_0$ , $\Delta f = f(x) - f(x_0)$ it becomes the first row
$\ln x \approx x - 1$	Natural log at $x_0 = 1$	$f(1) = \ln 1 = 0$ and $f'(1) = \tfrac{1}{1} = 1$

Key Insight: The two sides agree exactly at $x = x_0$ and stay close only while $x$ is near $x_0$ — equivalently, while $\Delta x$ is small. Far from the point of tangency nothing ties the curve to the line.

IIVisualization 1 — $\ln x \approx x - 1$ Only Near $x = 1$

The tangent line $y = x - 1$ equals $\ln x$ at $x = 1$ and separates from it as $x$ moves away.

Point

x

2.00

💡 How close to $x_0$ is close enough depends on the function and on the accuracy you need; a sharper answer comes later in this unit.

IIIVisualization 2 — Secant Slope Approaches $f'(x_0)$

As $\Delta x \to 0$ , the secant slope $\frac{\Delta f}{\Delta x}$ of $\ln x$ from $(1, 0)$ approaches the tangent slope $f'(1) = 1$ .

Shrink

\Delta x

2.000

IVVisualization 3 — From $\frac{\Delta f}{\Delta x} \approx f'(x_0)$ to the Tangent-Line Formula

Multiplying by $\Delta x$ and writing out the deltas turns the difference-quotient form into $f(x) \approx f(x_0) + f'(x_0)(x - x_0)$ .

Step 1

Step 1 — Difference quotient $\approx$ derivative

\frac{\Delta f}{\Delta x} \approx f'(x_0)

\frac{\sqrt{2.25} - \sqrt{1}}{1.25} = \frac{0.5}{1.25} = 0.4 \quad\text{vs}\quad f'(1) = \frac{1}{2\sqrt{1}} = 0.5

Step 2 — Multiply both sides by $\Delta x$

\Delta f \approx f'(x_0)\,\Delta x

0.5 \quad\text{vs}\quad 0.5 \times 1.25 = 0.625

Step 3 — Write out each delta

\Delta x = x - x_0, \qquad \Delta f = f(x) - f(x_0)

f(x) - f(x_0) \approx f'(x_0)(x - x_0)

\sqrt{x} - 1 \approx \tfrac{1}{2}(x - 1)

Step 4 — Add $f(x_0)$ to both sides

f(x) \approx f(x_0) + f'(x_0)(x - x_0)

\sqrt{x} \approx 1 + \tfrac{1}{2}(x - 1), \qquad \sqrt{2.25} \approx 1.625\ \ (\text{exact: } 1.5)

VQuiz Questions

Problem 1 · Estimate a Square Root

Given: $f(x) = \sqrt{x}$ with base point $x_0 = 9$ . Find the linear approximation of $\sqrt{9.2}$ .

✅ Correct!

3 + \frac{1}{6}(0.2) = 3 + \frac{1}{30} \approx 3.0333

, and the true value is

\sqrt{9.2} \approx 3.0332

❌ Check the slope.

f'(x) = \frac{1}{2\sqrt{x}}

must be evaluated at

x_0 = 9

f'(9) = \frac{1}{2 \cdot 3} = \frac{1}{6}

, not

\frac{1}{2}

❌ Missing a factor of 2. The derivative of

\sqrt{x} = x^{1/2}

\frac{1}{2}x^{-1/2}

, so

f'(9) = \frac{1}{6}

, not

\frac{1}{3}

❌ That uses slope $1$ .

\Delta f \approx \Delta x

holds for

\ln x

x_0 = 1

, not here: the slope of

\sqrt{x}

9

\frac{1}{6}

❌ Not quite. Use

f(9) = 3

f'(9) = \frac{1}{6}

and

x - x_0 = 0.2

Show solution

The two numbers at the base point: $f(9) = \sqrt{9} = 3$ and, since $f'(x) = \frac{1}{2\sqrt{x}}$ , $f'(9) = \frac{1}{6}$ .

\sqrt{x} \approx 3 + \tfrac{1}{6}(x - 9)

\sqrt{9.2} \approx 3 + \tfrac{1}{6}(0.2) = 3 + \tfrac{1}{30} \approx 3.0333

The exact value is $3.03315\ldots$ , so the error is under $0.0002$ — the step $0.2$ is small relative to the curvature of $\sqrt{x}$ near $9$ .

Problem 2 · Where Is the Derivative Evaluated?

Given: $f(x) = x^3$ with base point $x_0 = 1$ . Find the value the linear approximation gives for $(1.1)^3$ .

✅ Correct!

1 + 3(0.1) = 1.3

; the exact

1.331

0.031

higher because

x^3

bends upward away from its tangent line.

❌ That is the exact value.

(1.1)^3 = 1.331

, but the linear approximation is the tangent line:

1 + 3(0.1) = 1.3

❌ The slope must come from the base point. You used

f'(1.1) = 3.63

. In

f(x_0) + f'(x_0)(x - x_0)

the derivative is evaluated at

x_0 = 1

, giving the constant slope

3

— otherwise the right side is not a line.

❌ Missing the slope.

f'(x) = 3x^2

, so

f'(1) = 3

multiplies the step:

3 \times 0.1 = 0.3

Show solution

At the base point: $f(1) = 1$ and $f'(x) = 3x^2$ gives $f'(1) = 3$ . The base point goes in three places; $x$ stays the variable:

x^3 \approx f(1) + f'(1)(x - 1) = 1 + 3(x - 1)

(1.1)^3 \approx 1 + 3(0.1) = 1.3

Evaluating $f'$ at $1.1$ instead gives $1 + 3.63(0.1) = 1.363$ , which is not the tangent line at $x_0 = 1$ . The true value $1.331$ lies between the two.

Problem 3 · The Natural Log at a New Base Point

Given: $f(x) = \ln x$ with base point $x_0 = e$ . Find the linear approximation $L(x)$ , then use it to estimate $\ln 3$ .

What is the linear approximation near $x_0 = e$ ?

What estimate does it give for $\ln 3$ ?

✅ Correct!

L(3) = 3/e \approx 1.104

, against the true

\ln 3 \approx 1.0986

— close, because

3

is near the base point

e \approx 2.718

❌ Check the slope.

f'(x) = \frac{1}{x}

, so

f'(e) = \frac{1}{e}

, not

e

❌ That is the approximation at base point $1$ . At

x_0 = e

you need

f(e) = 1

and

f'(e) = \frac{1}{e}

❌ The value term is missing.

f(e) = \ln e = 1

, so the approximation starts at

1

1 + \frac{x - e}{e}

❌ That comes from $\ln x \approx x - 1$ . That line is tangent at

x = 1

, and

3

is far from

1

: it overshoots by almost

0.9

❌ Check $L(x)$ . Use

f(e) = 1

and

f'(e) = \frac{1}{e}

f(e) + f'(e)(x - e)

, then simplify.

❌ Check the estimate. Substitute

x = 3

into

L(x) = \frac{x}{e}

\frac{3}{2.71828\ldots}

Show solution

At $x_0 = e$ we know the logarithm exactly: $f(e) = \ln e = 1$ , and $f'(e) = \frac{1}{e}$ .

\ln x \approx 1 + \frac{1}{e}(x - e) = 1 + \frac{x}{e} - 1 = \frac{x}{e}

\ln 3 \approx \frac{3}{e} \approx 1.1036

The true value is $\ln 3 \approx 1.0986$ , an error of about $0.005$ . The base-point- $1$ formula gives $3 - 1 = 2$ , off by about $0.9$ : the same function, but $3$ is near $e$ and far from $1$ .

Problem 4 · Change in Area

Given: a circle's area is $A(r) = \pi r^2$ , and its radius grows from $r = 10$ to $r = 10.1$ . Find the estimate $\Delta A \approx A'(r)\,\Delta r$ at $r = 10$ .

✅ Correct!

A'(10)\,\Delta r = 20\pi \times 0.1 = 2\pi \approx 6.28

; the exact change is

2.01\pi

❌ That is only the rate.

A'(10) = 20\pi

is area per unit of radius; multiply by the step

\Delta r = 0.1

❌ That is the exact change.

\pi(10.1^2 - 10^2) = 2.01\pi

; the linear estimate drops the

\pi(\Delta r)^2

term and gives

2\pi

❌ That is $\pi(\Delta r)^2$ . That is the small leftover term the linear estimate ignores, not the estimate itself.

Show solution

Here the function is $A(r)$ , the base point is $r_0 = 10$ and $\Delta r = 0.1$ . With $A'(r) = 2\pi r$ , $A'(10) = 20\pi$ .

\Delta A \approx A'(10)\,\Delta r = 20\pi \times 0.1 = 2\pi \approx 6.28

Exact: $\Delta A = \pi\big((10.1)^2 - 10^2\big) = \pi(2 \cdot 10 \cdot 0.1 + 0.1^2) = 2.01\pi$ . The linear estimate keeps the $2 r_0\,\Delta r$ term and drops $(\Delta r)^2$ , which is tiny when $\Delta r$ is small.

Solved: 0 / 4

Linear Approximation: The Tangent Line as an Approximation

IKey Formulas

IIVisualization 1 — ln⁡x≈x−1\ln x \approx x - 1lnx≈x−1 Only Near x=1x = 1x=1

IIIVisualization 2 — Secant Slope Approaches f′(x0)f'(x_0)f′(x0​)

IVVisualization 3 — From ΔfΔx≈f′(x0)\frac{\Delta f}{\Delta x} \approx f'(x_0)ΔxΔf​≈f′(x0​) to the Tangent-Line Formula

VQuiz Questions

IIVisualization 1 — $\ln x \approx x - 1$ Only Near $x = 1$

IIIVisualization 2 — Secant Slope Approaches $f'(x_0)$

IVVisualization 3 — From $\frac{\Delta f}{\Delta x} \approx f'(x_0)$ to the Tangent-Line Formula