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

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

For a function that is differentiable at a point, its graph can be closely approximated by its tangent line near that point. If one were to "zoom in" infinitely on the point of tangency, the curve of the function would become indistinguishable from its tangent line. This property is known as local linearity. It allows us to use a simple linear function—the tangent line—as a substitute for a more complex function to estimate its values at points very close to the point of tangency.

The tangent line provides the best possible linear approximation of the function at that specific point because it matches the function's value and its instantaneous rate of change (the derivative) precisely at that location. This process of using the tangent line to approximate function values is called linearization.

Key Formulas/Rules/Theorems

The Equation of the Tangent Line and Linearization

The approximation of a function $f(x)$ for values of $x$ near a point $x=a$ is based on the equation of the tangent line to the graph of $f$ at $x=a$.

The equation for the tangent line is given in point-slope form as:

$$y-y_1=m(x-x_1)$$

At the point $(a,f(a))$, the slope $m$ is the derivative $f^{\prime }(a)$. Substituting these gives:

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

Solving for $y$ yields the equation used for linear approximation:

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

This equation defines the linearization of $f$ at $x=a$, which is denoted by $L(x)$.

The Linearization of $f$ at $a$:

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

For values of $x$ close to $a$, we can use the approximation $f(x)\approx L(x)$.

Understanding Local Linearity

The concept of local linearity is the foundation for why linearization works. A function is considered locally linear at a point $x=a$ if it is differentiable at that point. The tangent line is the "best" linear approximation for several reasons:

1. Shared Point: The tangent line passes through the point $(a,f(a))$, so the function and its linearization have the exact same value at $x=a$. That is, $L(a)=f(a)$.

2. Shared Slope: The slope of the tangent line is $f^{\prime }(a)$, which is the same as the instantaneous rate of change of the function $f$ at $x=a$.

No other straight line shares both these properties with the function $f$ at the point $x=a$. This dual agreement in value and rate of change ensures that the line "hugs" the curve most closely near that point.

It is critical to understand that the accuracy of the approximation $f(x)\approx L(x)$ diminishes as $x$ moves further away from $a$. The approximation is only reliable for values of $x$ in a small neighborhood around $a$.

Core Concepts & Rules

• Differentiability Implies Local Linearity: A function that is differentiable at a point $x=a$ can be approximated by a linear function (its tangent line) near that point.

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

• Required Components: To construct a linear approximation, three pieces of information are necessary:

  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, $f^{\prime }(a)$.

• Approximation: The value of the function $f(x)$ can be approximated by the value of its linearization $L(x)$ for $x$-values close to $a$. We write this as $f(x)\approx L(x)$.

Step-by-Step Example 1: Finding a Linear Approximation

Problem: Let $f(x)=\cos (x)$. Use the linearization of $f(x)$ at $x=\frac{\pi }{2}$ to approximate the value of $\cos (1.5)$.

Step 1: Identify the function and the center of approximation.

The function is $f(x)=\cos (x)$.

The center of approximation is $a=\frac{\pi }{2}$. We are using this point because $1.5$ is close to $\frac{\pi }{2}\approx 1.57$.

Step 2: Find the necessary components for the linearization formula $L(x)=f(a)+f^{\prime }(a)(x-a)$.

• Find $f(a)$:

  $f(\frac{\pi }{2})=\cos (\frac{\pi }{2})=0$

• Find the derivative, $f^{\prime }(x)$:

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

• Find $f^{\prime }(a)$:

  $f^{\prime }(\frac{\pi }{2})=-\sin (\frac{\pi }{2})=-1$

Step 3: Construct the linearization $L(x)$.

Substitute the values from Step 2 into the formula:

$L(x)=f(\frac{\pi }{2})+f^{\prime }(\frac{\pi }{2})(x-\frac{\pi }{2})$

$L(x)=0+(-1)(x-\frac{\pi }{2})$

$L(x)=-(x-\frac{\pi }{2})$

Step 4: Use $L(x)$ to approximate the desired value.

We want to approximate $\cos (1.5)$, so we evaluate $L(1.5)$:

$\cos (1.5)\approx L(1.5)=-(1.5-\frac{\pi }{2})$

$L(1.5)=\frac{\pi }{2}-1.5$

Using the approximation $\pi \approx 3.14159$:

$L(1.5)\approx \frac{3.14159}{2}-1.5=1.5708-1.5=0.0708$

Therefore, the linearization of $\cos (x)$ at $x=\frac{\pi }{2}$ approximates $\cos (1.5)$ to be about $0.0708$.

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

Problem: A function $f$ is twice-differentiable. The table below gives values of $f$ and its derivative $f^{\prime }$ for selected values of $x$. Use the tangent line to the graph of $f$ at $x=3$ to approximate $f(3.1)$.

Step 1: Identify the goal and the center of approximation.

The goal is to approximate $f(3.1)$.

The problem specifies using the tangent line at $x=3$, so the center of approximation is $a=3$.

Step 2: Find the necessary components from the table.

The linearization formula is $L(x)=f(a)+f^{\prime }(a)(x-a)$. We need $f(3)$ and $f^{\prime }(3)$.

• From the table, $f(3)=-2$.

• From the table, $f^{\prime }(3)=4$.

Step 3: Write the equation of the tangent line (linearization).

Substitute the values into the formula:

$L(x)=f(3)+f^{\prime }(3)(x-3)$

$L(x)=-2+4(x-3)$

Step 4: Use the tangent line to approximate the value.

To approximate $f(3.1)$, we calculate $L(3.1)$:

$f(3.1)\approx L(3.1)=-2+4(3.1-3)$

$L(3.1)=-2+4(0.1)$

$L(3.1)=-2+0.4$

$L(3.1)=-1.6$

The approximate value of $f(3.1)$ is $-1.6$.

Using Your Calculator

This topic is primarily analytical, and you will be expected to construct and use linearizations by hand. A calculator is not used to find the linearization itself, but it can be a powerful tool for verifying your result and understanding the concept.

To check your work and visualize the approximation:

1. Graph the Function: In your calculator's graphing menu, enter the original function into $Y_1$. For Example 1, $Y_1=\cos (X)$.

2. Graph the Linearization: Enter the linearization equation you found into $Y_2$. For Example 1, $Y_2=-(X-\pi /2)$.

3. Visualize the Tangency: Graph both functions. Use the ZOOM feature to zoom in on the point of tangency ($x=\pi /2$ in this case). You should see the line $Y_2$ appearing to be tangent to the curve $Y_1$ at that point. The closer you zoom, the more the curve will resemble the line.

4. Compare Values: Use the calculator's table feature or calculate values directly.

   • Calculate the actual value: Find $Y_1(1.5)$. The calculator will return $\cos (1.5)\approx 0.0707$.

   • Calculate the approximate value: Find $Y_2(1.5)$. The calculator will return $-(1.5-\pi /2)\approx 0.0708$.

   • This confirms that your linearization provides a very close approximation.

AP Exam Quick Hit

Common Question Types

• Given an Explicit Function: You will be given a function like $f(x)=\sqrt[3]{x}$ and asked to write the equation of the tangent line at $x=8$ to approximate $f(8.1)$. This is a direct test of the formula and your ability to compute derivatives.

• Given Tabular Data: A table of values for $f(x)$ and $f^{\prime }(x)$ will be provided. You will be asked to use the data to find the tangent line approximation at a specific $x$-value in the table, similar to Example 2 above. This tests your ability to extract the correct information ($f(a)$ and $f^{\prime }(a)$) and apply the formula.

• Given a Differential Equation: You might be given a differential equation like $\frac{dy}{dx}=2x-y$ and an initial condition, such as $y(1)=3$. You would then be asked to use the tangent line at $x=1$ to approximate $y(1.2)$. Here, $a=1$, $f(a)=3$, and you find $f^{\prime }(a)$ by plugging the point $(1,3)$ into the differential equation: $f^{\prime }(1)=2(1)-3=-1$.

Common Mistakes

• Using $x$ Instead of $a$: A frequent error is writing the linearization as $L(x)=f(x)+f^{\prime }(x)(x-a)$. The function value $f(a)$ and the slope $f^{\prime }(a)$ must be constant numerical values evaluated at the center of approximation, $a$.

• Incorrect Point of Tangency: Using the $x$-value you are approximating instead of the center $a$ in the formula. For example, when approximating $f(3.1)$ with a tangent at $a=3$, a student might incorrectly write $L(x)=f(3)+f^{\prime }(3)(x-3.1)$. The term must be $(x-a)$, which is $(x-3)$.

• Calculation Errors: Simple mistakes in calculating the derivative $f^{\prime }(x)$ or evaluating $f(a)$ and $f^{\prime }(a)$ can lead to an incorrect final answer, even if the correct process is followed.

• Stopping Too Soon: Students correctly find the equation of the tangent line, $L(x)$, but then fail to use it to find the requested approximation. Always re-read the prompt to ensure you have answered the full question, which usually involves plugging a value into your $L(x)$ equation.