Wemazi

Topic 4.6

Estimate a nearby function value

Builds on building a tangent-line approximation. 7 minutes.

A cube root without a calculator

The cube root of 28 is hard to find by hand, but the cube root of 27 is easy: it is 3.

The curve is \(f(x) = \sqrt[3]{x}\), and the orange point is \((27, 3)\). The slope of the orange tangent line is \(f'(27) = \tfrac{1}{27}\), because \(f'(x) = \dfrac{1}{3x^{2/3}}\) and \(27^{2/3} = 9\).

The line is \(L(x) = 3 + \tfrac{1}{27}(x - 27)\). Move the target to \(x = 28\). The orange dot on the line is at \(3 + \tfrac{1}{27}\), which is about 3.0370.

The chalk dot on the curve is at about 3.0366, so the two dots almost coincide. The height of the line at the target is the estimate of \(\sqrt[3]{28}\).

Estimate by adding a change

Drag the target \(x\). The faint horizontal segment from \(a\) to \(x\) is the offset \(x - a\), which is negative when the target is to the left of \(a\).

The orange vertical segment is the rise of the tangent line over the offset, \(f'(a)(x - a)\), which is the estimated change in \(f\).

For every target, the estimate is \(f(a)\) plus this change. The red segment between the line and the curve is the error, which is small when the target is near \(a\).

Notation

  1. Choose a center \(a\) near the target. The center is the input where the tangent line touches the curve, and both \(f(a)\) and \(f'(a)\) must be easy to find.

    \[a = 27, \; f(a) = 3, \; \cy{f'(a)} = \tfrac{1}{27}\]

  2. The offset is the target minus the center.

    \[\Delta x = x - a = 28 - 27 = 1\]

  3. Multiply the slope by the offset to find the estimated change in \(f\).

    \[\Delta f \approx \cy{f'(a)}\,\Delta x = \tfrac{1}{27}\]

  4. Add the estimated change to \(f(a)\).

    \[f(x) \approx f(a) + f'(a)\,\Delta x = 3 + \tfrac{1}{27}\]

    The estimate works when the offset is small, because the line uses only the value and the slope at a.

When it fails

Move the target to \(x = 64\), far from the center.

The line gives \(3 + \tfrac{37}{27}\), about 4.37, but \(\sqrt[3]{64} = 4\). The red segment between the line and the curve is long, because the curve bends away from the line over a large offset.

Example 1: A power

Use the tangent line to \(f(x) = x^5\) at \(a = 1\) to estimate \(1.03^5\).

  1. Find the value and the slope at the center.

    \[f(1) = 1, \qquad f'(1) = 5 \cdot 1^4 = 5\]

  2. The offset is \(1.03 - 1 = 0.03\). The estimated change is the slope times the offset.

    \[\Delta f \approx 5 \cdot 0.03 = 0.15\]

  3. Add the change to the value at the center.

    \[1.03^5 \approx 1 + 0.15 = \cy{1.15}\]

    As a check, \(1.03^5 \approx 1.1593\). The chalk dot for this value is almost under the orange dot.

Example 2: A square root

Use a tangent line to \(f(x) = \sqrt{x}\) to estimate \(\sqrt{10}\).

  1. The chalk dot shows the true height of the curve at the target \(x = 10\), which is about 3.16.

  2. A tempting wrong move: Choose \(a = 1\), because \(\sqrt{1} = 1\) is easy. The slope is \(f'(1) = \tfrac12\).

    \[L(10) = 1 + \tfrac12(10 - 1) = 5.5\]

    The curve is only about 3.16 high at \(x = 10\), but the red line reaches 5.5. The target is 9 units from the center, which is too far for the line to stay close to the curve.

  3. Choose \(a = 9\), because it is close to 10 and \(\sqrt{9} = 3\) is easy. The slope comes from \(f'(x) = \dfrac{1}{2\sqrt{x}}\).

    \[f(9) = 3, \qquad f'(9) = \tfrac{1}{2 \cdot 3} = \tfrac16\]

  4. The offset is \(10 - 9 = 1\), so the estimate is \(f(9)\) plus the slope.

    \[\sqrt{10} \approx 3 + \tfrac16 \cdot 1 = \cy{\tfrac{19}{6}}\]

    As a check, \(19/6 \approx 3.1667\) and \(\sqrt{10} \approx 3.1623\).

Practice

Use the tangent line to \(f(x) = x^{1/4}\) at \(a = 16\) to estimate \(16.2^{1/4}\).

The offset is the distance from 16 to 16.2. Multiply it by the slope at 16.

Solution

  1. The value at the center is the fourth root of 16. The slope is \(f'(x) = \tfrac14 x^{-3/4}\), so \(f'(16) = \tfrac14 \cdot \tfrac18\).

    \[f(16) = 2, \qquad f'(16) = \tfrac{1}{32}\]

  2. The offset is \(16.2 - 16 = 0.2\). The estimated change is the slope times the offset.

    \[\Delta f \approx \tfrac{1}{32} \cdot 0.2 = \tfrac{1}{160}\]

  3. Add the change to \(f(16)\).

    \[16.2^{1/4} \approx 2 + \tfrac{1}{160} = 2.00625\]

Questions

Questions 1 and 2 use the curve of \(k(x) = \tfrac{x^2}{2} + 3\) and its tangent line at \(x = 2\). The curve passes through \((2, 5)\), and the tangent line has slope 2.

  1. What is the tangent line estimate of \(k(2.1)\)?

    • 5.2
    • 7.1
    • 5.1
    Answer

    5.2

  2. Which target gives the better estimate, \(x = 2.1\) or \(x = 2.8\)?

    • \(x = 2.1\), because it is closer to the center
    • \(x = 2.8\), because the line is higher there
    • Both are equally good, because the same line is used
    Answer

    \(x = 2.1\), because it is closer to the center

  3. Use the tangent line to \(g(x) = e^{2x}\) at \(a = 0\) to estimate \(g(0.05)\).

    • 1.1
    • 1.05
    • 1.0
    Answer

    1.1

Summary

You choose a center \(a\) near the target, and you add the slope times the offset to \(f(a)\).

This works because near \(a\) the tangent line stays close to the curve, so its height is close to \(f(x)\).

You use it when you can find \(f(a)\) and \(f'(a)\) easily and the target is near \(a\).

This is one of four skills in topic 4.6 of Which & Why: AP Calculus.