Introduction

This topic focuses on using tangent lines to approximate function values near known points. The essential knowledge tells us that the tangent line is the graph of a locally linear approximation of the function near the point of tangency and for a tangent line approximation, the function's behavior near the point of tangency may determine whether a tangent line value is an underestimate or an overestimate of the corresponding function value. This powerful technique allows us to estimate values without computing them directly.

Local Linearity and Linearization

The Concept of Local Linearity

Near any point where a function is differentiable, the function looks approximately like a straight line when you "zoom in" close enough. This is called local linearity.

The tangent line at a point provides the best linear approximation to the function near that point.

The Linearization Formula

For a function $f(x)$ at point $x=a$, the linearization or tangent line approximation is:

$L(x)=f(a)+f^{\prime }(a)(x-a)$

This is the equation of the tangent line at $x=a$, written in point-slope form.

Components:

Example 1:

Find the linearization of $f(x)=x^2$ at $x=3$.

Find $f(3)$: $f(3)=9$

Find $f^{\prime }(x)$ and $f^{\prime }(3)$: $f^{\prime }(x)=2x$

$f^{\prime }(3)=6$

Linearization: $L(x)=9+6(x-3)$

$L(x)=9+6x-18$

$L(x)=6x-9$

Using Linearization to Approximate Values

Making Approximations

To approximate $f(x_0)$ for a value $x_0$ close to $a$, use $L(x_0)\approx f(x_0)$.

The approximation is most accurate when $x_0$ is very close to $a$.

Example 2:

Use the linearization from Example 1 to approximate $f(3.1)=(3.1{)}^2$.

$L(3.1)=6(3.1)-9$

$L(3.1)=18.6-9$

$L(3.1)=9.6$

Actual value: $(3.1{)}^2=9.61$

The approximation is very close!

Square Root Approximations

Example 3:

Approximate $\sqrt{26}$ using linearization at $x=25$.

Let $f(x)=\sqrt{x}$

Find $f(25)$: $f(25)=5$

Find $f^{\prime }(x)$ and $f^{\prime }(25)$: $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$

$f^{\prime }(25)=\frac{1}{2(5)}=\frac{1}{10}$

Linearization: $L(x)=5+\frac{1}{10}(x-25)$

Approximate $\sqrt{26}$: $L(26)=5+\frac{1}{10}(26-25)$

$L(26)=5+\frac{1}{10}$

$L(26)=5.1$

Actual value: $\sqrt{26}\approx 5.099$

Trigonometric Approximations

Example 4:

Approximate $\sin (0.1)$ using linearization at $x=0$.

Let $f(x)=\sin (x)$

Find $f(0)$: $f(0)=0$

Find $f^{\prime }(x)$ and $f^{\prime }(0)$: $f^{\prime }(x)=\cos (x)$

$f^{\prime }(0)=1$

Linearization: $L(x)=0+1(x-0)$

$L(x)=x$

Approximate $\sin (0.1)$: $L(0.1)=0.1$

Actual value: $\sin (0.1)\approx 0.0998$

Note: Near $x=0$, $\sin (x)\approx x$ is a famous approximation!

Determining Overestimates and Underestimates

Using Concavity

The relationship between the tangent line and the function depends on concavity:

Concave Up ($f^{\prime \prime }(x)>0$):

Concave Down ($f^{\prime \prime }(x)<0$):

Visual Understanding

For a concave up function:

For a concave down function:

Example 5:

Is the linearization of $f(x)=x^2$ at $x=3$ an overestimate or underestimate for $f(3.1)$?

Find $f^{\prime \prime }(x)$: $f^{\prime }(x)=2x$

$f^{\prime \prime }(x)=2>0$

Since $f^{\prime \prime }(x)>0$, the function is concave up.

The tangent line is an underestimate.

Verification: $L(3.1)=9.6$ and $f(3.1)=9.61$

Indeed, $9.6<9.61$ ✓

Analyzing Different Functions

Example 6:

Determine if the linearization of $f(x)=\ln (x)$ at $x=1$ overestimates or underestimates $f(1.2)$.

Find $f^{\prime \prime }(x)$: $f^{\prime }(x)=\frac{1}{x}$

$f^{\prime \prime }(x)=-\frac{1}{x^2}$

At $x=1$: $f^{\prime \prime }(1)=-1<0$

Since $f^{\prime \prime }(x)<0$, the function is concave down.

The tangent line is an overestimate.

Find the linearization: $f(1)=0$

$f^{\prime }(1)=1$

$L(x)=0+1(x-1)=x-1$

$L(1.2)=0.2$

$f(1.2)=\ln (1.2)\approx 0.182$

Indeed, $0.2>0.182$ ✓

Applications of Linear Approximation

Estimating Change

Linearization can estimate how much a function changes: $\Delta f\approx f^{\prime }(a)\cdot \Delta x$

Example 7:

The radius of a sphere is measured as 10 cm with a possible error of 0.1 cm. Estimate the maximum error in the calculated volume.

$V=\frac{4}{3}\pi r^3$

$V^{\prime }(r)=4\pi r^2$

At $r=10$: $V^{\prime }(10)=4\pi (100)=400\pi$

Change in volume: $\Delta V\approx 400\pi (0.1)$

$\Delta V\approx 40\pi \approx 125.7$ cm³

Real-World Context

Example 8:

A company's profit function is $P(x)=-0.01x^2+50x-1000$ dollars when producing $x$ units. If production is currently 200 units, use linearization to estimate the profit when producing 205 units.

Find $P(200)$: $P(200)=-0.01(40000)+50(200)-1000$

$P(200)=-400+10000-1000$

$P(200)=8600$

Find $P^{\prime }(x)$ and $P^{\prime }(200)$: $P^{\prime }(x)=-0.02x+50$

$P^{\prime }(200)=-4+50=46$

Linearization: $L(x)=8600+46(x-200)$

Estimate $P(205)$: $L(205)=8600+46(5)$

$L(205)=8600+230$

$L(205)=8830$

The profit is approximately $8830 when producing 205 units.

Accuracy of Approximations

Distance from Point of Tangency

The accuracy of linear approximation depends on:

Example 9:

Compare approximations for $\sqrt{25.5}$ and $\sqrt{30}$ using linearization at $x=25$.

From Example 3: $L(x)=5+\frac{1}{10}(x-25)$

For $\sqrt{25.5}$: $L(25.5)=5+\frac{1}{10}(0.5)=5.05$

Actual: $\sqrt{25.5}\approx 5.050$

Error: $\approx 0.000$ (very small)

For $\sqrt{30}$: $L(30)=5+\frac{1}{10}(5)=5.5$

Actual: $\sqrt{30}\approx 5.477$

Error: $\approx 0.023$ (larger)

The approximation is better when closer to $x=25$.

Problem-Solving Strategy

Step-by-Step Approach

Key Formulas Summary

Linearization: $L(x)=f(a)+f^{\prime }(a)(x-a)$

Differential form: $dy=f^{\prime }(x)dx$ or $\Delta y\approx f^{\prime }(a)\Delta x$

Concavity test:

Practice Section