Approximating Values of a Function Using Local Linearity and Linearization - AP Calculus AB Study Guide

The Core Idea: Approximating Values of a Function Using Local Linearity and Linearization

The fundamental concept of local linearity is that for any function that is differentiable at a point, if you "zoom in" sufficiently close to that point on its graph, the curve will appear to be a straight line. This straight line is the function's tangent line at that point. Because the tangent line is so close to the function's graph near the point of tangency, it serves as an excellent linear approximation of the function in that small neighborhood.

We can leverage this property to estimate the values of a function that might be difficult to calculate directly. By finding the equation of the tangent line at a "nice" point $x=a$ (where we know the function's value and its derivative), we can use the simple linear equation of that line to approximate the function's value at a nearby point. This process is called linearization, and the tangent line equation itself is referred to as the linearization of the function at that point.

Key Formulas

The central formula for this topic is the equation of the tangent line to a function $f(x)$ at a point $x=a$, which is used as the linear approximation.

Equation of the Tangent Line (Point-Slope Form)

The equation of the line tangent to the graph of $y=f(x)$ at the point $(a,f(a))$ is given by:

$$y-f(a)=f^{\prime }(a)(x-a)$$

Rearranging this into function form gives the linearization.

Linearization of a Function

The linearization of a function $f(x)$ at $x=a$, denoted $L(x)$, is the function whose graph is the tangent line at $x=a$. It is the best linear approximation of $f(x)$ for values of $x$ near $a$.

$$L(x)=f(a)+f^{\prime }(a)(x-a)$$

For values of $x$ close to $a$, we can use the approximation:

$$f(x)\approx L(x)$$

Understanding Local Linearity

The ability to use a tangent line for approximation is fundamentally tied to the concept of differentiability. A function $f$ is differentiable at $x=a$ if its derivative $f^{\prime }(a)$ exists. The existence of the derivative means there is a well-defined, non-vertical tangent line at that point.

The principle of local linearity states that for a differentiable function, the graph of $y=f(x)$ and its tangent line at $x=a$ are nearly indistinguishable for $x$-values in a small interval around $a$. This is the justification for why the approximation $f(x)\approx L(x)$ is valid.

It is critical to understand that this is a local approximation. The accuracy of using $L(x)$ to estimate $f(x)$ decreases as $x$ moves further away from $a$. The tangent line only provides a good approximation in the immediate vicinity of the point of tangency.

Core Concepts & Rules

• Best Linear Approximation: The tangent line to the graph of $f(x)$ at $x=a$ provides the best possible linear approximation of the function for values of $x$ near $a$.

• Differentiability Implies Local Linearity: If a function is differentiable at a point, it is also locally linear at that point. This means its graph can be approximated by its tangent line there.

• The Linearization Function: The function $L(x)=f(a)+f^{\prime }(a)(x-a)$ is called the linearization of $f$ at $a$.

• The Approximation: To approximate the value of $f(x)$ for an $x$ near $a$, you calculate $L(x)$ instead. The core approximation is $f(x)\approx f(a)+f^{\prime }(a)(x-a)$.

• Required Information: To find the linearization and make an approximation, you need three pieces of information:

  1. The point of tangency, $a$.

  2. The value of the function at that point, $f(a)$.

  3. The value of the derivative at that point, $f^{\prime }(a)$.

Step-by-Step Example 1: Basic Application

Problem: Use the tangent line to the graph of $f(x)=\sqrt[3]{x}$ at $x=8$ to approximate the value of $\sqrt[3]{8.1}$.

Step 1: Identify the function and the point of tangency.

The function is $f(x)=x^{1/3}$.

The approximation is centered at the "nice" point $a=8$.

Step 2: Find the value of the function at $a=8$ (the y-coordinate).

$$f(8)=\sqrt[3]{8}=2$$

The point of tangency is $(8,2)$.

Step 3: Find the derivative of the function and evaluate it at $a=8$ (the slope).

First, find the derivative $f^{\prime }(x)$:

$$f^{\prime }(x)=\frac{d}{dx}(x^{1/3})=\frac{1}{3}{x}^{-2/3}=\frac{1}{3\sqrt[3]{x^2}}$$

Now, evaluate the derivative at $x=8$:

$$f^{\prime }(8)=\frac{1}{3\sqrt[3]{8^2}}=\frac{1}{3\sqrt[3]{64}}=\frac{1}{3(4)}=\frac{1}{12}$$

The slope of the tangent line at $x=8$ is $\frac{1}{12}$.

Step 4: Write the equation of the tangent line (the linearization $L(x)$).

Using the formula $L(x)=f(a)+f^{\prime }(a)(x-a)$:

$$L(x)=2+\frac{1}{12}(x-8)$$

Step 5: Use the linearization to approximate the desired value.

We want to approximate $f(8.1)=\sqrt[3]{8.1}$. We do this by calculating $L(8.1)$.

$$f(8.1)\approx L(8.1)=2+\frac{1}{12}(8.1-8)$$

$$L(8.1)=2+\frac{1}{12}(0.1)=2+\frac{1}{120}$$

$$L(8.1)=\frac{240}{120}+\frac{1}{120}=\frac{241}{120}$$

Therefore, the approximation is $\sqrt[3]{8.1}\approx \frac{241}{120}$.

Step-by-Step Example 2: Exam-Style Application

Problem: A function $f$ is twice-differentiable. Selected values of $f$ and its derivative $f^{\prime }$ are given in the table below. Use the equation of the line tangent to the graph of $f$ at $x=5$ to approximate the value of $f(4.9)$.

Step 1: Identify the point of tangency and the necessary values.

The problem specifies using the tangent line at $x=5$, so $a=5$.

From the table, we find the function value and the derivative value at $x=5$:

• $f(5)=-3$

• $f^{\prime }(5)=4$

Step 2: Write the equation of the tangent line at $x=5$ in point-slope form.

The point is $(a,f(a))=(5,-3)$.

The slope is $m=f^{\prime }(5)=4$.

The equation is:

$$y-(-3)=4(x-5)$$

$$y+3=4(x-5)$$

Step 3: Use the tangent line equation to approximate $f(4.9)$.

We substitute $x=4.9$ into our equation and solve for $y$. This $y$ value will be our approximation for $f(4.9)$.

$$y+3=4(4.9-5)$$

$$y+3=4(-0.1)$$

$$y+3=-0.4$$

$$y=-0.4-3$$

$$y=-3.4$$

Conclusion: The approximation for $f(4.9)$ is $-3.4$.

Using Your Calculator

The process of finding a linear approximation is primarily analytical and does not require a calculator. The steps involve finding a derivative and performing arithmetic, which are expected to be done by hand.

However, a calculator can be used in two main ways:

1. Finding the Derivative Value: If the function $f(x)$ is complex and you are asked to find the linearization at $x=a$, you can use the calculator to find the numerical derivative $f^{\prime }(a)$.

   • On a TI-84 style calculator, use the nDeriv command (found under MATH -> 8).

   • The syntax is nDeriv(function, X, value).

   • For Example 1, to find $f^{\prime }(8)$ for $f(x)=\sqrt[3]{x}‘,youwouldenter‘nDeriv(X^{(}1/3),X,8)‘.Thecalculatorwouldreturnavalueverycloseto$\frac{1}{12}(approx. 0.0833). 2. **Checking Your Answer:** After you have calculated your approximation $L(x), you can use the calculator to find the actual value of $f(x)$ and see how close your approximation is.

   • For Example 1, we approximated $\sqrt[3]{8.1}\approx \frac{241}{120}\approx 2.00833$.

   • Using a calculator, the actual value is $\sqrt[3]{8.1}\approx 2.008298$.

   • This confirms that our linear approximation is very close to the actual value.

AP Exam Quick Hit

Common Question Types

• Direct Calculation from a Function: You are given an explicit function like $f(x)=\sin (x)+x^2$, asked to find the equation of the tangent line at $x=0$, and then use it to approximate $f(0.1)$.

• Table-Based Problem: As in Example 2, you are provided a table of values for $f(x)$ and $f^{\prime }(x)$ and asked to use the data to build a tangent line at a specific point and approximate a nearby value. This is a very common format on both multiple-choice and free-response sections.

• Differential Equation Problem: You are given a differential equation $\frac{dy}{dx}=g(x,y)$ and an initial condition $(a,f(a))$. You are asked to use the tangent line at $(a,f(a))$ to approximate $f(b)$ for a $b$ near $a$. The slope $f^{\prime }(a)$ is found by plugging the coordinates $(a,f(a))$ into the $\frac{dy}{dx}$ expression.

Common Mistakes

• Using the Wrong Slope: Students incorrectly calculate the slope of the tangent line at the point they are approximating ($x=b$) instead of the point of tangency ($x=a$). The slope is constant and is always $f^{\prime }(a)$.

• Incorrect Point-Slope Formula: Mixing up the components of the tangent line equation, for example, writing $y-a=f^{\prime }(a)(x-f(a))$. Remember it is $y-y_1=m(x-x_1)$.

• Arithmetic Errors: Simple calculation mistakes when plugging the value into the tangent line equation, especially with negative signs or fractions.

• Conceptual Error: Forgetting the $(x-a)$ term entirely and just calculating $f(a)+f^{\prime }(a)$. The change in $x$ is critical to the approximation.

• Plugging $\Delta x$ into $f^{\prime }(a)$: Calculating $f(a)+f^{\prime }(a)\cdot \Delta x$ but accidentally plugging $x$ into $f^{\prime }(a)$ instead of $a$. The slope is fixed at the point of tangency.