AP Calculus BC 4.6 Approximating Values of a Function Using Local Linearity and Linearization Study Notes - New Syllabus
AP Calculus BC 4.6 Approximating Values of a Function Using Local Linearity and Linearization Study Notes- New syllabus
AP Calculus BC 4.6 Approximating Values of a Function Using Local Linearity and Linearization Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Approximate a value on a curve using the equation of a tangent line.
Key Concepts:
- Approximating Values of a Function Using Local Linearity
- Linearization of a Function at a Point
Approximating Values of a Function Using Local Linearity and Linearization
Approximating Values of a Function Using Local Linearity
When a function is differentiable at a point, it behaves almost like a straight line (linear) very close to that point. This idea is known as local linearity.
Near a point \( x = a \), the value of the function \( f(x) \) can be approximated using the tangent line at that point.
Formula (Tangent Line Approximation):
If \( f(x) \) is differentiable at \( x = a \), then for values of \( x \) close to \( a \),
\( f(x) \approx f(a) + f'(a)(x – a) \)
Example:
Approximate \( \sqrt{4.1} \) using local linearity.
▶️Answer/Explanation
Let \( f(x) = \sqrt{x} \). Choose \( a = 4 \) since it’s easy to compute.
- \( f(4) = 2 \)
- \( f'(x) = \dfrac{1}{2\sqrt{x}} \), so \( f'(4) = \dfrac{1}{4} \)
Now approximate:
\( f(4.1) \approx f(4) + f'(4)(4.1 – 4) = 2 + \frac{1}{4}(0.1) = 2.025 \)
Approximate value: \( \sqrt{4.1} \approx 2.025 \)
Example:
Approximate \( \ln(1.05) \) using local linearity.
▶️Answer/Explanation
Let \( f(x) = \ln x \). Choose \( a = 1 \).
- \( f(1) = \ln 1 = 0 \)
- \( f'(x) = \dfrac{1}{x} \), so \( f'(1) = 1 \)
Now approximate:
\( f(1.05) \approx 0 + 1(1.05 – 1) = 0.05 \)
Approximate value: \( \ln(1.05) \approx 0.05 \)
Linearization of a Function at a Point
Linearization is the process of creating a linear function that approximates the behavior of a function near a specific point. The linearization of \( f(x) \) at \( x = a \) is essentially the equation of the tangent line at that point.
Linearization Formula:
\( L(x) = f(a) + f'(a)(x – a) \)
This function \( L(x) \) is called the linearization of \( f \) at \( x = a \).
Example:
Find the linearization of \( f(x) = \cos x \) at \( x = 0 \).
▶️Answer/Explanation
- \( f(0) = \cos 0 = 1 \)
- \( f'(x) = -\sin x \), so \( f'(0) = 0 \)
Then:
\( L(x) = 1 + 0(x – 0) = 1 \)
Linearization: \( L(x) = 1 \)
Example:
Find the linearization of \( f(x) = e^x \) at \( x = 0 \)
▶️Answer/Explanation
- \( f(0) = e^0 = 1 \)
- \( f'(x) = e^x \), so \( f'(0) = 1 \)
Then:
\( L(x) = 1 + 1(x – 0) = 1 + x \)
Linearization: \( L(x) = 1 + x \)
Approximating a Value on a Curve Using the Tangent Line
To approximate the value of a function \( f(x) \) near a point \( x = a \), you can use the tangent line at that point.
The tangent line gives a linear approximation to the curve and is especially useful when calculating the actual value is difficult.
Formula:
\( f(x) \approx f(a) + f'(a)(x – a) \)
This is the equation of the tangent line at \( x = a \) used to estimate \( f(x) \) for values close to \( a \).
Example:
Use the tangent line to \( f(x) = \sqrt{x} \) at \( x = 9 \) to approximate \( \sqrt{9.2} \).
▶️Answer/Explanation
- \( f(x) = \sqrt{x} \)
- Let \( a = 9 \). Then \( f(9) = \sqrt{9} = 3 \)
- \( f'(x) = \dfrac{1}{2\sqrt{x}} \)
- \( f'(9) = \dfrac{1}{2 \cdot 3} = \dfrac{1}{6} \)
Use the tangent line formula to approximate
\( f(x) \approx f(a) + f'(a)(x – a) \) \( f(9.2) \approx 3 + \frac{1}{6}(9.2 – 9) = 3 + \frac{1}{6}(0.2) = 3 + \frac{0.2}{6} = 3 + 0.0333 = 3.0333 \)
Conclusion: \( \sqrt{9.2} \approx 3.0333 \) using the tangent line.
Example :
Let \( f \) be the function given by \( f(x) = 3x^2 – 4x + 2 \). The tangent line to the graph of \( f \) at \( x = 1 \) is used to approximate values of \( f(x) \).
Which of the following is the smallest value of \( x \) for which the error resulting from this tangent line approximation is more than 0.5?
- 1.3
- 1.4
- 1.5
- 1.6
- 1.7
▶️Answer/Explanation
- \( f(x) = 3x^2 – 4x + 2 \)
- \( f(1) = 3(1)^2 – 4(1) + 2 = 3 – 4 + 2 = 1 \)
- \( f'(x) = 6x – 4 \)
- \( f'(1) = 6(1) – 4 = 2 \)
Tangent line (linearization) formula
The tangent line at \( x = 1 \) is: \( L(x) = f(1) + f'(1)(x – 1) = 1 + 2(x – 1) \)
Compute the approximation and actual value for each option and check the error
Try \( x = 1.3 \):
- \( L(1.3) = 1 + 2(1.3 – 1) = 1 + 0.6 = 1.6 \)
- \( f(1.3) = 3(1.3)^2 – 4(1.3) + 2 = 3(1.69) – 5.2 + 2 = 5.07 – 5.2 + 2 = 1.87 \)
- \( |1.87 – 1.6| = 0.27 \) → Not > 0.5
Try \( x = 1.4 \):
- \( L(1.4) = 1 + 2(0.4) = 1.8 \)
- \( f(1.4) = 3(1.96) – 5.6 + 2 = 5.88 – 5.6 + 2 = 2.28 \)
- \( |2.28 – 1.8| = 0.48 \) → Still not > 0.5
Try \( x = 1.5 \):
- \( L(1.5) = 1 + 2(0.5) = 2 \)
- \( f(1.5) = 3(2.25) – 6 + 2 = 6.75 – 6 + 2 = 2.75 \)
- \( |2.75 – 2| = 0.75 \) → Error > 0.5
Smallest x for which error > 0.5 is 1.5
Correct Answer: (C) 1.5