Local Linearity and Linearization - AP Calculus AB

  1. State the linear approximation for f(x)=1xf(x) = \frac{1}{x} at x=2x = 2.
    L(x)=12−14(x−2)L(x) = \frac{1}{2} - \frac{1}{4}(x - 2). Since f(2)=12f(2) = \frac{1}{2} and f′(2)=−14f'(2) = -\frac{1}{4}.
  2. Which function approximates f(x)f(x) near x=ax = a using local linearity?
    The linearization L(x)L(x) of f(x)f(x). The best linear approximation near the given point.
  3. What is the linear approximation for f(x)=sqrt(x)f(x) = \text{sqrt}(x) at x=4x = 4?
    L(x)=2+14(x−4)L(x) = 2 + \frac{1}{4}(x - 4). Since 4=2\sqrt{4} = 2 and (x)′=12x(\sqrt{x})' = \frac{1}{2\sqrt{x}}.
  4. Determine the linearization of f(x)=x2f(x) = x^2 at x=2x = 2.
    L(x)=4+4(x−2)L(x) = 4 + 4(x - 2). Since f(2)=4f(2) = 4 and f′(x)=2xf'(x) = 2x, so f′(2)=4f'(2) = 4.
  5. What is the linear approximation of f(x)=x4f(x) = x^4 at x=1x = 1?
    L(x)=1+4(x−1)L(x) = 1 + 4(x - 1). Since f(1)=1f(1) = 1 and f′(x)=4x3f'(x) = 4x^3, so f′(1)=4f'(1) = 4.
  6. State the condition under which a function is locally linear at a point.
    The derivative f′(x)f'(x) exists and is continuous near the point. Ensures the function behaves like a line near that point.
  7. Determine the linear approximation of f(x)=x3f(x) = x^3 at x=1x = 1.
    L(x)=1+3(x−1)L(x) = 1 + 3(x - 1). Since f(1)=1f(1) = 1 and f′(x)=3x2f'(x) = 3x^2, so f′(1)=3f'(1) = 3.
  8. Find the linear approximation of f(x)=sin(x)f(x) = \text{sin}(x) at x=0x = 0.
    L(x)=xL(x) = x. Since sin⁡(0)=0\sin(0) = 0 and sin⁡′(0)=cos⁡(0)=1\sin'(0) = \cos(0) = 1.
  9. Calculate the derivative needed for linearization of f(x)=1xf(x) = \frac{1}{x} at x=1x = 1.
    f′(x)=−1x2f'(x) = -\frac{1}{x^2}, so f′(1)=−1f'(1) = -1. Using the power rule and evaluating at x=1x = 1.
  10. Identify the linear approximation for f(x)=exp(x)f(x) = \text{exp}(x) at x=1x = 1.
    L(x)=e+e(x−1)L(x) = e + e(x - 1). Since e1=ee^1 = e and (ex)′=ex(e^x)' = e^x, so f′(1)=ef'(1) = e.
  11. Which function value is used in the linear approximation L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a)?
    f(a)f(a). The y-intercept when the tangent line passes through (a,f(a))(a, f(a)).
  12. What is the formula for the linear approximation of a function at point aa?
    L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a). This is the tangent line equation at point aa.
  13. What is the linear approximation of f(x)=exf(x) = e^x at x=0x = 0?
    L(x)=1+xL(x) = 1 + x. Since e0=1e^0 = 1 and (ex)′=ex(e^x)' = e^x, so f′(0)=1f'(0) = 1.
  14. What role does f′(a)f'(a) play in linearization?
    It is the slope of the tangent line at x=ax = a. It determines the steepness of the linear approximation.
  15. Find the linear approximation of f(x)=arcsin(x)f(x) = \text{arcsin}(x) at x=0x = 0.
    L(x)=xL(x) = x. Since arcsin⁡(0)=0\arcsin(0) = 0 and (arcsin⁡x)′=11−x2(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}.
  16. What is the approximation L(x)L(x) when f(x)=tan(x)f(x) = \text{tan}(x) at x=0x = 0?
    L(x)=xL(x) = x. Since tan⁡(0)=0\tan(0) = 0 and tan⁡′(0)=sec⁡2(0)=1\tan'(0) = \sec^2(0) = 1.
  17. What is the linear approximation of f(x)=x5f(x) = x^5 at x=1x = 1?
    L(x)=1+5(x−1)L(x) = 1 + 5(x - 1). Since f(1)=1f(1) = 1 and f′(x)=5x4f'(x) = 5x^4, so f′(1)=5f'(1) = 5.
  18. Which point is used as the center for a linear approximation?
    The point aa where the approximation is centered. The base point where the tangent line touches the curve.
  19. Identify the derivative required for linearization of f(x)f(x) at x=ax = a.
    f′(a)f'(a). The slope of the tangent line at the linearization point.
  20. What is the linear approximation of f(x)=ln(x)f(x) = \text{ln}(x) at x=1x = 1?
    L(x)=x−1L(x) = x - 1. Since ln⁡(1)=0\ln(1) = 0 and (ln⁡x)′=1x(\ln x)' = \frac{1}{x}, so f′(1)=1f'(1) = 1.
  21. What is the linear approximation of f(x)=ln(x)f(x) = \text{ln}(x) at x=ex = e?
    L(x)=1+1e(x−e)L(x) = 1 + \frac{1}{e}(x - e). Since ln⁡(e)=1\ln(e) = 1 and (ln⁡x)′=1x(\ln x)' = \frac{1}{x}, so f′(e)=1ef'(e) = \frac{1}{e}.
  22. What is the linear approximation of f(x)=arctan(x)f(x) = \text{arctan}(x) at x=0x = 0?
    L(x)=xL(x) = x. Since arctan⁡(0)=0\arctan(0) = 0 and (arctan⁡x)′=11+x2(\arctan x)' = \frac{1}{1+x^2}.
  23. Identify the linear approximation of f(x)=cos(x)f(x) = \text{cos}(x) at x=0x = 0.
    L(x)=1L(x) = 1. Since cos⁡(0)=1\cos(0) = 1 and cos⁡′(0)=−sin⁡(0)=0\cos'(0) = -\sin(0) = 0.

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