AcademyApplications Of Derivatives
Academy
Linear Approximation
Level 1 - Calculus topic page in Applications Of Derivatives.
Principle
Linear approximation replaces a differentiable function near a point by its tangent line. It is useful when the tangent line is easier to compute than the original function.
Notation
\(L(x)\)
linear approximation to f near a
\(a\)
base point
\(f'(a)\)
slope of the tangent line at a
The Core Method
Use the tangent line at \(x=a\).
Linear approximation
\[L(x)=f(a)+f'(a)(x-a)\]
The approximation is best when \(x\) is close to \(a\). The farther \(x\) is from \(a\), the more curvature can affect the error.
Worked Cases
Question
Use a linear approximation to estimate \(\sqrt{4.1}\).
Answer
Let \(f(x)=\sqrt{x}\) and choose \(a=4\). Then \(f(4)=2\), and \(f'(x)=1/(2\sqrt{x})\), so \(f'(4)=1/4\). Thus \(L(x)=2+\frac14(x-4)\). At \(x=4.1\), \(L(4.1)=2+0.025=2.025\).
Examples
Question
Approximate \((2.02)^3\) using \(f(x)=x^3\) near \(a=2\).
Answer
Here \(f(2)=8\) and \(f'(x)=3x^2\), so \(f'(2)=12\). The linear approximation is \(L(x)=8+12(x-2)\). At \(x=2.02\), \(L=8+12(0.02)=8.24\).