Local Linearity and Linearization - AP Calculus BC

1. Evaluate $L(x)$ for $f(x) = \text{cos}(x)$ at $x = \frac{\text{π}}{2}$ using $x = \frac{\text{π}}{2} + 0.1$.

   $L(x) = 0 - 0.1 = -0.1$. $f(\frac{\pi}{2}) = 0$, $f'(\frac{\pi}{2}) = -1$, so slope is negative.

2. Which function value does $f(a)$ represent in the linear approximation formula?

   The value of the function at $x = a$. This is the y-coordinate where the tangent line touches the curve.

3. Determine the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$ for $x = e+0.1$.

   $L(x) = 1 + 0.1/e$. $f(e) = 1$, $f'(e) = \frac{1}{e}$, so slope is $\frac{1}{e}$.

4. What is the formula for the linear approximation of a function $f$ at $x=a$?

   $L(x) = f(a) + f'(a)(x-a)$. Standard formula where $f(a)$ is the point and $f'(a)$ is the slope.

5. Define local linearity in the context of a differentiable function.

   Local linearity means the function appears linear near a point. The function's graph looks like a straight line when zoomed in close.

6. Determine the linear approximation of $f(x) = x^3$ at $x = 1$ for $x = 1.05$.

   $L(x) = 1 + 3(0.05) = 1.15$. $f(1) = 1$, $f'(1) = 3$, so $L(1.05) = 1 + 3(0.05)$.

7. Evaluate $L(x)$ for $f(x) = \frac{1}{x}$ at $x = 2$ using $x = 2.1$.

   $L(x) = \frac{1}{2} - \frac{0.1}{4} = 0.475$. $f(2) = 0.5$, $f'(2) = -0.25$, so $L(2.1) = 0.5 - 0.25(0.1)$.

8. Evaluate $L(x)$ for $f(x) = \text{sqrt}(x)$ at $x = 4$ using $x = 4.1$.

   $L(x) = 2 + \frac{0.1}{4} = 2.025$. $f(4) = 2$, $f'(4) = \frac{1}{4}$, so $L(4.1) = 2 + \frac{1}{4}(0.1)$.

9. Calculate the linear approximation of $f(x) = x^2$ at $x=1$ for $x=1.1$.

   $1 + 2(0.1) = 1.2$. $f(1) = 1$, $f'(1) = 2$, so $L(1.1) = 1 + 2(0.1)$.

10. What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?

    $L(x) = x - 1$. $f(1) = 0$, $f'(1) = 1$, giving the simple linear form.

11. Approximate $e^{0.2}$ using linearization at $x = 0$ for $f(x) = e^x$.

    $L(x) = 1 + 0.2 = 1.2$. $f(0) = 1$, $f'(0) = 1$, so $L(0.2) = 1 + 1(0.2)$.

12. What does $f'(a)(x-a)$ represent in the linearization formula?

    The change in the linear approximation. This represents how much the function changes linearly.

13. Calculate the linear approximation of $f(x) = x^3$ at $x = 0$ for $x = 0.1$.

    $L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.

14. Determine $L(x)$ for $f(x) = e^x$ at $x = 0$ with $x = 0.05$.

    $L(x) = 1 + 0.05 = 1.05$. $f(0) = 1$, $f'(0) = 1$, so $L(0.05) = 1 + 1(0.05)$.

15. What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$?

    $L(x) = 1 + \frac{x-e}{e}$. $f(e) = 1$, $f'(e) = \frac{1}{e}$, standard logarithm linearization.

16. What is the approximated change in $f(x)$ for $f(x) = x^3$ at $x = 1$?

    $3(x-1)$. For $f(x) = x^3$ at $x=1$, $f'(1) = 3$ gives the linear change.

17. What is the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$?

    For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = 1$, so linearization is just $x$.

18. What is the role of the derivative in linear approximation?

    It gives the slope of the tangent line. The derivative provides the rate of change for the approximation.

19. What is $L(x)$ in the linear approximation formula?

    The linear approximation of $f(x)$ near $x = a$. The linearization function that approximates $f(x)$.

20. What is the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$?

    For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = \sec^2(0) = 1$, so linearization is $x$.

21. Approximate $e^{0.1}$ using linearization at $x = 0$ for $f(x) = e^x$.

    $L(x) = 1 + 0.1 = 1.1$. $f(0) = 1$, $f'(0) = 1$, so $L(0.1) = 1 + 1(0.1)$.

22. For which type of function can local linearity be used to approximate values?

    Differentiable functions. Must have a derivative to calculate the tangent line slope.

23. What is the linear approximation of $f(x) = \frac{1}{x}$ at $x = 3$?

    $L(x) = \frac{1}{3} - \frac{1}{9}(x-3)$. $f(3) = \frac{1}{3}$, $f'(3) = -\frac{1}{9}$, standard linearization formula.

24. Approximate $\text{sin}(0.1)$ using linearization at $x = 0$ for $f(x) = \text{sin}(x)$.

    $L(x) = 0.1$. $f(0) = 0$, $f'(0) = 1$, so $L(0.1) = 0 + 1(0.1)$.

25. Find the linear approximation of $f(x) = x^2$ at $x = 0$ for $x = 0.1$.

    $L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.

26. Determine the linear approximation of $f(x) = x^4$ at $x = 1$ for $x = 1.1$.

    $L(x) = 1 + 0.4 = 1.4$. $f(1) = 1$, $f'(1) = 4$, so $L(1.1) = 1 + 4(0.1)$.

27. What is the linear approximation of $f(x) = e^x$ at $x = 1$?

    $L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, standard exponential linearization.

28. What is the geometric interpretation of a linearization at $x = a$?

    It's the tangent line to the function at $x = a$. The linearization creates the tangent line at that specific point.

29. Approximate $\text{ln}(1.1)$ using linearization at $x = 1$ for $f(x) = \text{ln}(x)$.

    $L(x) = 0 + 0.1 = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.

30. What is the linear approximation of $f(x) = \text{exp}(x)$ at $x = 1$?

    $L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, so linearization has slope $e$.

31. Find $L(x)$ for $f(x) = \text{ln}(x)$ at $x = 1$ when $x = 1.1$.

    $L(x) = 0 + \frac{1}{1}(0.1) = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.

32. Identify the derivative in the linearization formula $L(x) = f(a) + f'(a)(x-a)$.

    $f'(a)$. This term represents the slope of the tangent line.

33. Find the linear approximation of $f(x) = \frac{1}{x}$ at $x = 1$ for $x = 0.9$.

    $L(x) = 1 + 0.1 = 1.1$. $f(1) = 1$, $f'(1) = -1$, so $L(0.9) = 1 + (-1)(-0.1)$.

34. Find the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$ for $x = 0.05$.

    $L(x) = 0.05$. $f(0) = 0$, $f'(0) = 1$, so $L(0.05) = 0 + 1(0.05)$.

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