4.6 Approximating Values of a Function Using Local Linearity and Linearization

Learning objectives

—CHA-3.F—Approximate a value on a curve using the equation of a tangent line• CHA-3.F Approximate a value on a curve using the equation of a tangent line.• CHA-3.F.1 The tangent line is the graph of a locally linear approximation of the function near the point of tangency.• CHA-3.F.2 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.• Enduring understanding CHA-3: Derivatives allow us to solve real-world problems involving rates of change.

Near $x=a$ , a differentiable function is approximately linear, so its tangent line can stand in for the curve. The linearization uses the known value $f(a)$ and the tangent slope $f'(a)$ .

L(x)=f(a)+f'(a)(x-a),\qquad f(x)\approx L(x)\text{ for }x\text{ near }a

Choose a nearby base point $a$ where $f(a)$ and $f'(a)$ are easy to evaluate.\n2. Build $L(x)=f(a)+f'(a)(x-a)$ .\n3. Substitute the target input into $L$ .\n4. Use local concavity to classify the estimate.

To approximate $\sqrt{4.1}$ , let $f(x)=\sqrt{x}$ and use $a=4$ . Since $f(4)=2$ and $f'(4)=1/4$ , $L(x)=2+\frac14(x-4).$ Therefore $\sqrt{4.1}\approx L(4.1)=2+\frac14(0.1)=2.025.$

Where the curve is concave up, it lies locally above its tangent line, so $L(x)$ is an underestimate. Where the curve is concave down, it lies locally below its tangent line, so $L(x)$ is an overestimate. Because $\sqrt{x}$ is concave down near $4$ , $2.025$ is an overestimate.

The approximation is not an identity: $f(x)$ and $L(x)$ agree in value and slope at $x=a$ , but their difference can grow as $x$ moves farther from the point of tangency.

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—4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate FormsRead