Unit 4.6 – Approximating Values of a Function Using Local Linearity and Linearization

AP® Calculus AB & BC | Tangent Line Approximations & Error Bounds

Local linearity means every smooth function looks like its tangent line when you zoom in close. Linearization uses this property to estimate function values using the tangent line equation, which is especially useful for values near a point where we can compute the function's value exactly.

🔑 The Linearization Formula

Linearization (Tangent Line Approximation)

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

Here, \(L(x)\) approximates \(f(x)\) for values of \(x\) near \(a\). This is simply the equation of the tangent line to the graph of \(f\) at the point \(x=a\)!

🔍 Using Local Linearity for Estimation

\(f(a)\): The actual function value at a known point (the "point of tangency" or "anchor").
\(f'(a)\): The slope of the function at that same point (the slope of the tangent line).
To estimate \(f(x)\) for a value of \(x\) close to \(a\), you plug it into \(L(x)\).
This is also known as the first-order Taylor polynomial centered at \(x=a\).

📚 Worked Examples

Example 1: Approximating \(\sqrt{25.2}\)

Let \(f(x) = \sqrt{x}\) and choose a nearby "nice" point, \(a=25\).
We have \(f(25)=5\). The derivative is \(f'(x)=\frac{1}{2\sqrt{x}}\), so \(f'(25)=\frac{1}{2\sqrt{25}} = 0.1\).
The linearization is: \(L(x) = 5 + 0.1(x-25)\).
Now, approximate at \(x=25.2\): \(L(25.2) = 5 + 0.1(25.2 - 25) = 5 + 0.1(0.2) = 5.02\). So, \(\sqrt{25.2} \approx 5.02\).

Example 2: Approximating \(e^{0.03}\)

Let \(f(x) = e^x\) and choose the anchor point \(a=0\).
We have \(f(0)=1\) and \(f'(x)=e^x\), so \(f'(0)=1\).
The linearization is \(L(x) = 1 + 1(x-0) = 1 + x\).
So, \(L(0.03)=1+0.03=1.03\). Thus, \(e^{0.03} \approx 1.03\).

💡 Linearization Tips, Tricks & Short Notes

Linear approximations are most accurate for values of \(x\) very close to the anchor point \(a\).
The farther \(x\) is from \(a\), the less accurate the approximation \(L(x)\) becomes.
Linearization will underestimate the true value when the function is concave up (\(f''>0\)) and overestimate when the function is concave down (\(f''<0\)).
This is a very useful technique for estimating "weird" roots or transcendental function values near "nice" numbers.
Graphically, the linearization is the tangent line that "hugs" the curve at \(x=a\).

📝 Practice Problems

Try These Yourself:

Approximate \(\sin(0.12)\) using the linearization of \(f(x)=\sin(x)\) at \(x=0\).
Estimate \(\ln(1.05)\) using the tangent line to \(f(x)=\ln(x)\) at \(x=1\).
Use linearization at \(x=16\) to estimate the value of \(\sqrt{16.5}\).

Model Answers:

For \(f(x)=\sin(x)\) at \(a=0\): \(f(0)=0\), \(f'(0)=\cos(0)=1\). So \(L(x)=0+1(x-0)=x\). Thus, \(L(0.12)=0.12\).
For \(f(x)=\ln(x)\) at \(a=1\): \(f(1)=0\), \(f'(1)=1/1=1\). So \(L(x)=0+1(x-1)=x-1\). Thus, \(L(1.05)=0.05\).
For \(f(x)=\sqrt{x}\) at \(a=16\): \(f(16)=4\), \(f'(16)=1/(2\sqrt{16})=1/8\). So \(L(x)=4 + \frac{1}{8}(x-16)\). Thus, \(L(16.5)=4 + \frac{1}{8}(0.5) = 4 + 0.0625 = 4.0625\).

✏️ AP® Exam Success: Linearization

Always clearly state the function \(f(x)\), the anchor point \(a\), the values of \(f(a)\) and \(f'(a)\), and the resulting linearization formula \(L(x)\).
Show your differentiation and substitution steps clearly.
The units of \(L(x)\) and \(f(x)\) are the same.
Frame your answer in the context of the problem: “The value of \(f(x)\) can be approximated by its tangent line at \(x=a\).”
If asked whether the approximation is an over- or underestimate, use the sign of the second derivative (\(f''\)) to justify your conclusion based on concavity.
Box your final numerical answer for clarity.