Local Linearity and Linearization

Zoom in on a differentiable function near a point, and its graph looks more and more like its tangent line. So near the point, the tangent line's values are good estimates of the function's values.

From a formula

Example 1

Use the linearization of

f(x) = \sqrt[3]{x}

x = 64

to estimate

\sqrt[3]{62}

, and determine whether the estimate is too large or too small.

f(64) = 4

and

f'(x) = \frac{1}{3}x^{-2/3}

, so

f'(64) = \frac{1}{3} \cdot \frac{1}{16} = \frac{1}{48}

. Here

x - a = 62 - 64 = -2

\begin{aligned} \sqrt[3]{62} &\approx 4 + \frac{1}{48}(-2) \\[4pt] &= 4 - \frac{1}{24} \approx 3.958. \end{aligned}

Since

f''(x) = -\frac{2}{9}x^{-5/3} < 0

for

x > 0

, the graph is concave down, and the estimate is too large: an overestimate. The true value is

3.95789\ldots

From a graph of the derivative

Example 2

The graph of

f'

is shown for

0 \le x \le 2

, and

f(1) = 3

. Use the line tangent to the graph of

f

x = 1

to estimate

f(1.1)

. Determine whether the estimate is an overestimate or an underestimate, and explain.

The graph of the derivative f′, passing through (1, 2) and increasing on 0 ≤ x ≤ 2.

From the graph,

f'(1) = 2

, so

\begin{aligned} f(1.1) &\approx f(1) + f'(1)(1.1 - 1) \\[4pt] &= 3 + 2(0.1) = 3.2. \end{aligned}

The graph of

f'

is increasing on

1 \le x \le 1.1

, so the graph of

f

is concave up there. The tangent line lies below the graph of

f

, so

3.2

is an underestimate.

A curve given implicitly

Example 3

The curve

y^3 + xy = 14

passes through

(3, 2)

, and near that point it's the graph of a function

y = f(x)

. (a) Estimate

f(3.6)

with the tangent line at

x = 3

. (b) Find

\frac{d^2y}{dx^2}

, and use it to decide whether the estimate is too large or too small.

(a) Implicitly,

3y^2\,\frac{dy}{dx} + y + x\,\frac{dy}{dx} = 0

, so

\frac{dy}{dx} = -\frac{y}{3y^2 + x}.

(3, 2)

the slope is

-\frac{2}{15}

, and

f(3.6) \approx 2 - \frac{2}{15}(0.6) = 1.92

(b) Differentiating again and substituting

\frac{dy}{dx}

gives

\frac{d^2y}{dx^2} = \frac{2xy}{\left(3y^2 + x\right)^3}.

For

x > 0

on the curve,

y

can't be

0

or negative: then

y^3 + xy

would be at most

0

and couldn't equal

14

. So for

3 \le x \le 3.6

\frac{d^2y}{dx^2} > 0

: the graph is concave up, and

1.92

is an underestimate.

Proving an inequality

Running the approximation backward

Example 5

A cake cools on a counter. Its temperature

H(t)

, in degrees Fahrenheit, satisfies

H(10) = 150

and

H'(10) = -4

, with

t

in minutes. For

t \ge 10

H

is decreasing and its graph is concave up, and the cake cools to

140

degrees at some time

T > 10

. (a) Use the tangent line at

t = 10

to estimate when the cake reaches

140

degrees. (b) Determine whether the cake reaches

140

degrees before or after that estimate, and explain.

(a) Set the tangent line equal to

140

and solve for

t

150 - 4(t - 10) = 140.

That gives

t = 12.5

minutes.

(b) The graph is concave up, so it lies above the tangent line. At

t = 12.5

the tangent line is at

140

, so

H(12.5) > 140

. Since

H

is decreasing,

H(t) > 140

for every

t

from

10

12.5

. So

T > 12.5

: the cake reaches

140

degrees after the estimate.