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The derivative of the function f is defined by f^(')(x)=(x^(2)-4x)sin(3x+2). If f(3)=-6, then use a calculator to find the value of f(-2) to the nearest thousandth. Answer:

The derivative of the function f f is defined by f′(x)=(x2−4x)sin⁡(3x+2) f^{\prime}(x)=\left(x^{2}-4 x\right) \sin (3 x+2) . If f(3)=−6 f(3)=-6 , then use a calculator to find the value of f(−2) f(-2) to the nearest thousandth.\newlineAnswer:

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Q. The derivative of the function f f is defined by f′(x)=(x2−4x)sin⁡(3x+2) f^{\prime}(x)=\left(x^{2}-4 x\right) \sin (3 x+2) . If f(3)=−6 f(3)=-6 , then use a calculator to find the value of f(−2) f(-2) to the nearest thousandth.\newlineAnswer:
  1. Integrate f′(x)f'(x): To find f(−2)f(-2), we need to integrate f′(x)f'(x) to get f(x)f(x) and then use the initial condition f(3)=−6f(3) = -6 to find the constant of integration.\newlineIntegrate f′(x)=(x2−4x)sin⁡(3x+2)f'(x) = (x^2 - 4x)\sin(3x + 2).\newlineThis requires integration by parts or a special technique since it is a product of a polynomial and a trigonometric function.
  2. Apply integration by parts: Let's set u=x2−4xu = x^2 - 4x and dv=sin⁡(3x+2)dxdv = \sin(3x + 2)dx. Then we need to find dudu and vv.\newlineDifferentiate uu to get du=(2x−4)dxdu = (2x - 4)dx.\newlineIntegrate dvdv to get v=−13cos⁡(3x+2)v = -\frac{1}{3} \cos(3x + 2).
  3. Simplify and split integral: Apply the integration by parts formula: ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du. Substitute uu, dudu, vv, and dvdv into the formula. f(x)=−(x2−4x)(13cos⁡(3x+2))−∫(−13cos⁡(3x+2))(2x−4) dxf(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) - \int(-\frac{1}{3} \cos(3x + 2))(2x - 4)\,dx.
  4. Integrate by parts again: Simplify the integral and split it into two separate integrals. f(x)=−(x2−4x)(13cos⁡(3x+2))+(23)∫(xcos⁡(3x+2))dx−(43)∫cos⁡(3x+2)dxf(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{3})\int(x \cos(3x + 2))dx - (\frac{4}{3})\int\cos(3x + 2)dx.
  5. Integrate remaining integral: The first integral, ∫(xcos⁡(3x+2)) dx\int(x \cos(3x + 2))\,dx, again requires integration by parts. Let's set u=xu = x and dv=cos⁡(3x+2)dxdv = \cos(3x + 2)dx. Differentiate uu to get du=dxdu = dx. Integrate dvdv to get v=13sin⁡(3x+2)v = \frac{1}{3} \sin(3x + 2).
  6. Combine all parts: Apply the integration by parts formula to the first integral.\newlineThis gives us (13)xsin⁡(3x+2)−∫(13sin⁡(3x+2)) dx(\frac{1}{3})x \sin(3x + 2) - \int(\frac{1}{3} \sin(3x + 2))\,dx.
  7. Simplify and integrate: The remaining integral, ∫sin⁡(3x+2) dx\int \sin(3x + 2)\,dx, is straightforward. Integrate to get −13cos⁡(3x+2)-\frac{1}{3} \cos(3x + 2).
  8. Use initial condition: Combine all parts to express f(x)f(x).f(x)=−(x2−4x)(13cos⁡(3x+2))+(23)(13xsin⁡(3x+2)−∫13sin⁡(3x+2) dx)−(43)(−13cos⁡(3x+2)).f(x) = -\left(x^2 - 4x\right)\left(\frac{1}{3} \cos(3x + 2)\right) + \left(\frac{2}{3}\right)\left(\frac{1}{3}x \sin(3x + 2) - \int \frac{1}{3} \sin(3x + 2)\,dx\right) - \left(\frac{4}{3}\right)\left(-\frac{1}{3} \cos(3x + 2)\right).
  9. Calculate constant of integration: Simplify the expression for f(x)f(x) and integrate the remaining term.f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)+(19)cos⁡(3x+2)−(49)cos⁡(3x+2).f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) + (\frac{1}{9}) \cos(3x + 2) - (\frac{4}{9}) \cos(3x + 2).
  10. Find f(−2)f(-2): Combine like terms to get the final expression for f(x)f(x).\newlinef(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2).f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2).
  11. Find f(−2)f(-2): Combine like terms to get the final expression for f(x)f(x).
    f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2)f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2).Use the initial condition f(3)=−6f(3) = -6 to find the constant of integration CC.
    Substitute x=3x = 3 into the expression for f(x)f(x) and solve for CC.
    −6=−(32−4⋅3)(13cos⁡(3⋅3+2))+(29)⋅3sin⁡(3⋅3+2)−(13)cos⁡(3⋅3+2)+C-6 = -(3^2 - 4\cdot3)(\frac{1}{3} \cos(3\cdot3 + 2)) + (\frac{2}{9})\cdot3 \sin(3\cdot3 + 2) - (\frac{1}{3}) \cos(3\cdot3 + 2) + C.
  12. Find f(−2)f(-2): Combine like terms to get the final expression for f(x)f(x).
    f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2)f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2).Use the initial condition f(3)=−6f(3) = -6 to find the constant of integration CC.
    Substitute x=3x = 3 into the expression for f(x)f(x) and solve for CC.
    −6=−(32−4⋅3)(13cos⁡(3⋅3+2))+(29)⋅3sin⁡(3⋅3+2)−(13)cos⁡(3⋅3+2)+C-6 = -(3^2 - 4\cdot3)(\frac{1}{3} \cos(3\cdot3 + 2)) + (\frac{2}{9})\cdot3 \sin(3\cdot3 + 2) - (\frac{1}{3}) \cos(3\cdot3 + 2) + C.
    Calculate the trigonometric values and solve for CC.
    f(x)f(x)00.
    f(x)f(x)11.
    f(x)f(x)22.
    f(x)f(x)33.
  13. Find f(−2)f(-2): Combine like terms to get the final expression for f(x)f(x).
    f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2)f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2).Use the initial condition f(3)=−6f(3) = -6 to find the constant of integration CC.
    Substitute x=3x = 3 into the expression for f(x)f(x) and solve for CC.
    −6=−(32−4⋅3)(13cos⁡(3⋅3+2))+(29)⋅3sin⁡(3⋅3+2)−(13)cos⁡(3⋅3+2)+C-6 = -(3^2 - 4\cdot3)(\frac{1}{3} \cos(3\cdot3 + 2)) + (\frac{2}{9})\cdot3 \sin(3\cdot3 + 2) - (\frac{1}{3}) \cos(3\cdot3 + 2) + C.
    Calculate the trigonometric values and solve for CC.
    f(x)f(x)00.
    f(x)f(x)11.
    f(x)f(x)22.
    f(x)f(x)33.
    Now that we have CC, we can find f(−2)f(-2) by substituting f(x)f(x)66 into the expression for f(x)f(x).
    f(x)f(x)88.
  14. Find f(−2)f(-2): Combine like terms to get the final expression for f(x)f(x).
    f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2)f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2).Use the initial condition f(3)=−6f(3) = -6 to find the constant of integration CC.
    Substitute x=3x = 3 into the expression for f(x)f(x) and solve for CC.
    −6=−(32−4⋅3)(13cos⁡(3⋅3+2))+(29)⋅3sin⁡(3⋅3+2)−(13)cos⁡(3⋅3+2)+C-6 = -(3^2 - 4\cdot3)(\frac{1}{3} \cos(3\cdot3 + 2)) + (\frac{2}{9})\cdot3 \sin(3\cdot3 + 2) - (\frac{1}{3}) \cos(3\cdot3 + 2) + C.
    Calculate the trigonometric values and solve for CC.
    f(x)f(x)00.
    f(x)f(x)11.
    f(x)f(x)22.
    f(x)f(x)33.
    Now that we have CC, we can find f(−2)f(-2) by substituting f(x)f(x)66 into the expression for f(x)f(x).
    f(x)f(x)88.
    Calculate the trigonometric values and simplify to find f(−2)f(-2).
    f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2)f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2)00.
  15. Find f(−2)f(-2): Combine like terms to get the final expression for f(x)f(x).
    f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2)f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2).Use the initial condition f(3)=−6f(3) = -6 to find the constant of integration CC.
    Substitute x=3x = 3 into the expression for f(x)f(x) and solve for CC.
    −6=−(32−4⋅3)(13cos⁡(3⋅3+2))+(29)⋅3sin⁡(3⋅3+2)−(13)cos⁡(3⋅3+2)+C-6 = -(3^2 - 4\cdot3)(\frac{1}{3} \cos(3\cdot3 + 2)) + (\frac{2}{9})\cdot3 \sin(3\cdot3 + 2) - (\frac{1}{3}) \cos(3\cdot3 + 2) + C.
    Calculate the trigonometric values and solve for CC.
    f(x)f(x)00.
    f(x)f(x)11.
    f(x)f(x)22.
    f(x)f(x)33.
    Now that we have CC, we can find f(−2)f(-2) by substituting f(x)f(x)66 into the expression for f(x)f(x).
    f(x)f(x)88.
    Calculate the trigonometric values and simplify to find f(−2)f(-2).
    f(x)=−(x2−4x)(13cos⁡(3x+2))+(29)xsin⁡(3x+2)−(13)cos⁡(3x+2)f(x) = -(x^2 - 4x)(\frac{1}{3} \cos(3x + 2)) + (\frac{2}{9})x \sin(3x + 2) - (\frac{1}{3}) \cos(3x + 2)00.
    Use a calculator to find the trigonometric values and compute f(−2)f(-2) to the nearest thousandth.
    This step requires a calculator, and the exact values will depend on the calculator's output.

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