Linear Approximation | Math Coach Amy

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Key Ideas Easy Medium Hard

\(L(x) = f(a) + f'(a)(x-a)\)

concave up: underestimate; concave down: overestimate

What to Remember

The tangent line is the approximation

\(L(x) = f(a) + f'(a)(x - a)\)

Use a nearby \(a\) where \(f(a)\) and \(f'(a)\) are easy.

Write the line before you substitute the nearby input

Find \(f(a)\) and \(f'(a)\) on their own lines. Build \(L(x)\). Then put in the nearby \(x\).

Concavity tells you which side the estimate falls on

If \(f''(x) > 0\) near \(a\), the graph is concave up and the tangent line sits below the curve. \(L(x)\) is an underestimate.

If \(f''(x) < 0\) near \(a\), \(L(x)\) is an overestimate.

A table can replace the formula

If you are given \(f(a)\) and \(f'(a)\), you already have the two numbers in \(L(x)\).

Level 1

Easy

Build L(x) at a friendly input, then step a small distance.

Example

Use the tangent line to \(f(x) = \sqrt{x}\) at \(a = 9\) to approximate \(\sqrt{9.2}\).

\(f(9) = 3\)

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

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

\(L(x) = 3 + \dfrac{1}{6}(x - 9)\)

\(L(9.2) = 3 + \dfrac{1}{6}(0.2)\)

\(3 + \dfrac{1}{30}\)

\(\dfrac{91}{30}\)

Answer: \(\dfrac{91}{30}\)

Approximate \(f(3.1)\) for \(f(x) = x^2\) using the tangent line at \(a = 3\).

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\(f(3) = 9\)

\(f'(x) = 2x\)

\(f'(3) = 6\)

\(L(3.1) = 9 + 6(0.1)\)

\(9.6\)

Answer: \(9.6\)

Approximate \(e^{0.1}\) using the tangent line to \(f(x) = e^x\) at \(a = 0\).

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\(f(0) = 1\)

\(f'(0) = 1\)

\(L(0.1) = 1 + 0.1\)

\(1.1\)

Answer: \(1.1\)

Approximate \(\sqrt[3]{8.3}\) using \(f(x) = x^{1/3}\) at \(a = 8\).

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\(f(8) = 2\)

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

\(f'(8) = \dfrac{1}{12}\)

\(L(8.3) = 2 + \dfrac{1}{12}(0.3)\)

\(2 + \dfrac{1}{40}\)

\(\dfrac{81}{40}\)

Answer: \(\dfrac{81}{40}\)

Approximate \(\ln 1.2\) using the tangent line to \(f(x) = \ln x\) at \(a = 1\).

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\(f(1) = 0\)

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

\(f'(1) = 1\)

\(L(1.2) = 0.2\)

Answer: \(0.2\)

For \(f(x) = x^2\), the approximation of \(f(3.1)\) from the tangent line at \(3\) is 9.6. Is that an overestimate or an underestimate?

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\(f''(x) = 2 > 0\)

The graph is concave up, so the tangent line lies below the curve.

9.6 is an underestimate.

Answer: underestimate

Level 2

Medium

A cubic, a reciprocal, or values from a table. Then name the concavity.

Example

For \(f(x) = x^3 - x\), find the tangent-line approximation of \(f(1.9)\) using \(a = 2\).

\(f(2) = 6\)

\(f'(x) = 3x^2 - 1\)

\(f'(2) = 11\)

\(L(x) = 6 + 11(x - 2)\)

\(L(1.9) = 6 + 11(-0.1)\)

\(4.9\)

Answer: \(4.9\)

Approximate \(\dfrac{1}{3.98}\) using \(f(x) = \dfrac{1}{x}\) at \(a = 4\).

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\(f(4) = \dfrac{1}{4}\)

\(f'(x) = -\dfrac{1}{x^2}\)

\(f'(4) = -\dfrac{1}{16}\)

\(L(3.98) = \dfrac{1}{4} + \left(-\dfrac{1}{16}\right)(-0.02)\)

\(\dfrac{1}{4} + \dfrac{1}{800}\)

\(\dfrac{201}{800}\)

Answer: \(\dfrac{201}{800}\)

You are told \(f(2) = 5\) and \(f'(2) = -3\). Estimate \(f(2.4)\).

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\(L(x) = 5 + (-3)(x - 2)\)

\(L(2.4) = 5 + (-3)(0.4)\)

\(5 - 1.2\)

\(3.8\)

Answer: \(3.8\)

For \(f(x) = \sqrt{x}\), is the approximation \(\dfrac{91}{30}\) of \(\sqrt{9.2}\) too high or too low?

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

\(f''(x) = -\dfrac{1}{4} x^{-3/2} < 0\) for \(x > 0\).

The graph is concave down, so the tangent line is above the curve.

The approximation is too high.

Answer: overestimate

For \(f(x) = e^x\), is 1.1 an overestimate or an underestimate of \(e^{0.1}\)?

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\(f''(x) = e^x > 0\)

The tangent line is below the curve.

1.1 is an underestimate.

Answer: underestimate

Which line is the linearization of \(f(x) = x^3 - x\) at \(x = 2\)?

A. \(L(x) = 6 + 11(x - 2)\)

B. \(L(x) = 6 + 6(x - 2)\)

C. \(L(x) = 8 + 11(x - 2)\)

D. \(L(x) = 11 + 6(x - 2)\)

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\(f(2) = 8 - 2 = 6\)

\(f'(2) = 12 - 1 = 11\)

\(L(x) = 6 + 11(x - 2)\)

Answer: A. \(L(x) = 6 + 11(x - 2)\)

Level 3

Hard

A root that is shifted, an inverse tangent, and a comparison of two approximations.

Example

Approximate \(\sqrt{4.1}\) using \(f(x) = \sqrt{x + 1}\) at \(a = 3\).

\(f(3) = 2\)

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

\(f'(3) = \dfrac{1}{4}\)

\(L(3.1) = 2 + \dfrac{1}{4}(0.1)\)

\(2.025\)

Answer: \(2.025\)

Is 2.025 an overestimate or an underestimate of \(\sqrt{4.1}\) from the tangent line to \(\sqrt{x + 1}\) at \(x = 3\)?

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\(f''(x) = -\dfrac{1}{4}(x + 1)^{-3/2} < 0\)

The graph is concave down.

The tangent approximation is an overestimate.

Answer: overestimate

Approximate \(\arctan 1.1\) using the tangent line to \(f(x) = \arctan x\) at \(a = 1\).

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\(f(1) = \dfrac{\pi}{4}\)

\(f'(x) = \dfrac{1}{1 + x^2}\)

\(f'(1) = \dfrac{1}{2}\)

\(L(1.1) = \dfrac{\pi}{4} + \dfrac{1}{2}(0.1)\)

\(\dfrac{\pi}{4} + 0.05\)

Answer: \(\dfrac{\pi}{4} + 0.05\)

For \(f(x) = \arctan x\) near \(x = 1\), is the tangent-line approximation an overestimate or an underestimate?

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\(f'(x) = (1 + x^2)^{-1}\)

\(f''(x) = -2x(1 + x^2)^{-2}\)

For \(x > 0\), \(f''(x) < 0\).

The approximation is an overestimate.

Answer: overestimate

Use \(f(x) = \sqrt{x}\) at \(a = 4\) to approximate both \(\sqrt{4.4}\) and \(\sqrt{3.6}\).

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\(f(4) = 2\)

\(f'(4) = \dfrac{1}{4}\)

\(L(4.4) = 2 + \dfrac{1}{4}(0.4) = 2.1\)

\(L(3.6) = 2 + \dfrac{1}{4}(-0.4) = 1.9\)

Answer: \(\sqrt{4.4}\approx 2.1\), \(\sqrt{3.6}\approx 1.9\)

\(f(4) = 9\), \(f'(4) = -2\), and \(f''(x) < 0\) for all \(x\) near 4. Which statement is true of the estimate \(f(4.5) \approx 8\)?

A. 8 is larger than \(f(4.5)\).

B. 8 is smaller than \(f(4.5)\).

C. 8 equals \(f(4.5)\).

D. The estimate cannot be formed from the given information.

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\(L(4.5) = 9 + (-2)(0.5) = 8\)

\(f''(x) < 0\), so the tangent line is above the curve.

8 is larger than \(f(4.5)\).

Answer: A. 8 is larger than \(f(4.5)\)

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