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The derivative of the function 
f is defined by 
f^(')(x)=(x^(2)-3x)cos(2x). If 
f(0)=5, then use a calculator to find the value of 
f(6) to the nearest thousandth.
Answer:

The derivative of the function f f is defined by f(x)=(x23x)cos(2x) f^{\prime}(x)=\left(x^{2}-3 x\right) \cos (2 x) . If f(0)=5 f(0)=5 , then use a calculator to find the value of f(6) f(6) to the nearest thousandth.\newlineAnswer:

Full solution

Q. The derivative of the function f f is defined by f(x)=(x23x)cos(2x) f^{\prime}(x)=\left(x^{2}-3 x\right) \cos (2 x) . If f(0)=5 f(0)=5 , then use a calculator to find the value of f(6) f(6) to the nearest thousandth.\newlineAnswer:
  1. Integrate f(x)f'(x) for f(x)f(x): To find f(6)f(6), we need to integrate the derivative f(x)f'(x) to get the original function f(x)f(x). We will then use the initial condition f(0)=5f(0) = 5 to find the constant of integration.
  2. Use Integration by Parts: The integral of f(x)=(x23x)cos(2x)f'(x) = (x^2 - 3x)\cos(2x) is not straightforward due to the product of a polynomial and a trigonometric function. We will use integration by parts or a computer algebra system to find the integral.
  3. Find dudu and vv: Let's denote u=x23xu = x^2 - 3x and dv=cos(2x)dxdv = \cos(2x)dx. Then we need to find dudu and vv: du=(2x3)dxdu = (2x - 3)dx and v=12sin(2x)v = \frac{1}{2}\sin(2x).
  4. Integrate by Parts Again: Applying integration by parts, we get:\newlinef(x)dx=udv=uvvdu\int f'(x)\,dx = \int u\, dv = uv - \int v\, du\newline= (x23x)(12)sin(2x)(12)sin(2x)(2x3)dx(x^2 - 3x)(\frac{1}{2})\sin(2x) - \int(\frac{1}{2})\sin(2x)(2x - 3)\,dx.
  5. Evaluate Initial Condition: We need to integrate (12)sin(2x)(2x3)dx(\frac{1}{2})\sin(2x)(2x - 3)\,dx, which again requires integration by parts or a computer algebra system.
  6. Calculate Constant of Integration: Let's denote u=2x3u = 2x - 3 and dv=(12)sin(2x)dxdv = (\frac{1}{2})\sin(2x)dx. Then we need to find dudu and vv: du=2dxdu = 2dx and v=(14)cos(2x)v = -(\frac{1}{4})\cos(2x).
  7. Final Function f(x)f(x): Applying integration by parts again, we get:\newline(12)sin(2x)(2x3)dx=uvvdu\int (\frac{1}{2})\sin(2x)(2x - 3)\,dx = uv - \int v\,du\newline= (14)(2x3)cos(2x)-(\frac{1}{4})(2x - 3)\cos(2x) - \int -(\frac{11}{44})\cos(22x)(22)\,dx.
  8. Evaluate f(6)f(6): The integral of 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx is 14(2)(12)sin(2x)=sin(2x)4-\frac{1}{4}(2)(\frac{1}{2})\sin(2x) = -\frac{\sin(2x)}{4}.
  9. Evaluate f(6)f(6): The integral of 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx is 14(2)12sin(2x)=sin(2x)4-\frac{1}{4}(2)\frac{1}{2}\sin(2x) = -\frac{\sin(2x)}{4}. Combining all parts, we get: f(x)=(x23x)12sin(2x)+14(2x3)cos(2x)+sin(2x)4+Cf(x) = (x^2 - 3x)\frac{1}{2}\sin(2x) + \frac{1}{4}(2x - 3)\cos(2x) + \frac{\sin(2x)}{4} + C, where CC is the constant of integration.
  10. Evaluate f(6)f(6): The integral of 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx is 14(2)12sin(2x)=sin(2x)4-\frac{1}{4}(2)\frac{1}{2}\sin(2x) = -\frac{\sin(2x)}{4}. Combining all parts, we get: f(x)=(x23x)12sin(2x)+14(2x3)cos(2x)+sin(2x)4+Cf(x) = (x^2 - 3x)\frac{1}{2}\sin(2x) + \frac{1}{4}(2x - 3)\cos(2x) + \frac{\sin(2x)}{4} + C, where CC is the constant of integration. Using the initial condition f(0)=5f(0) = 5, we find CC: f(0)=(0230)12sin(20)+14(203)cos(20)+sin(20)4+C=5f(0) = (0^2 - 3\cdot 0)\frac{1}{2}\sin(2\cdot 0) + \frac{1}{4}(2\cdot 0 - 3)\cos(2\cdot 0) + \frac{\sin(2\cdot 0)}{4} + C = 5. Since sin(0)=0\sin(0) = 0 and cos(0)=1\cos(0) = 1, this simplifies to: 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx00.
  11. Evaluate f(6)f(6): The integral of 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx is 14(2)12sin(2x)=sin(2x)4-\frac{1}{4}(2)\frac{1}{2}\sin(2x) = -\frac{\sin(2x)}{4}. Combining all parts, we get: f(x)=(x23x)12sin(2x)+14(2x3)cos(2x)+sin(2x)4+Cf(x) = (x^2 - 3x)\frac{1}{2}\sin(2x) + \frac{1}{4}(2x - 3)\cos(2x) + \frac{\sin(2x)}{4} + C, where CC is the constant of integration. Using the initial condition f(0)=5f(0) = 5, we find CC: f(0)=(0230)12sin(20)+14(203)cos(20)+sin(20)4+C=5f(0) = (0^2 - 3\cdot 0)\frac{1}{2}\sin(2\cdot 0) + \frac{1}{4}(2\cdot 0 - 3)\cos(2\cdot 0) + \frac{\sin(2\cdot 0)}{4} + C = 5. Since sin(0)=0\sin(0) = 0 and cos(0)=1\cos(0) = 1, this simplifies to: 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx00. Now we have the function 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx11.
  12. Evaluate f(6)f(6): The integral of 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx is 14(2)12sin(2x)=sin(2x)4-\frac{1}{4}(2)\frac{1}{2}\sin(2x) = -\frac{\sin(2x)}{4}. Combining all parts, we get: f(x)=(x23x)12sin(2x)+14(2x3)cos(2x)+sin(2x)4+C,f(x) = (x^2 - 3x)\frac{1}{2}\sin(2x) + \frac{1}{4}(2x - 3)\cos(2x) + \frac{\sin(2x)}{4} + C, where CC is the constant of integration. Using the initial condition f(0)=5f(0) = 5, we find CC: f(0)=(0230)12sin(20)+14(203)cos(20)+sin(20)4+C=5.f(0) = (0^2 - 3\cdot0)\frac{1}{2}\sin(2\cdot0) + \frac{1}{4}(2\cdot0 - 3)\cos(2\cdot0) + \frac{\sin(2\cdot0)}{4} + C = 5. Since sin(0)=0\sin(0) = 0 and cos(0)=1\cos(0) = 1, this simplifies to: C=5+34.C = 5 + \frac{3}{4}. Now we have the function f(x)=(x23x)12sin(2x)+14(2x3)cos(2x)+sin(2x)4+5+34.f(x) = (x^2 - 3x)\frac{1}{2}\sin(2x) + \frac{1}{4}(2x - 3)\cos(2x) + \frac{\sin(2x)}{4} + 5 + \frac{3}{4}. We can now evaluate f(6)f(6) using a calculator: f(6)=(6236)12sin(26)+14(263)cos(26)+sin(26)4+5+34.f(6) = (6^2 - 3\cdot6)\frac{1}{2}\sin(2\cdot6) + \frac{1}{4}(2\cdot6 - 3)\cos(2\cdot6) + \frac{\sin(2\cdot6)}{4} + 5 + \frac{3}{4}.
  13. Evaluate f(6)f(6): The integral of 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx is 14(2)12sin(2x)=sin(2x)4-\frac{1}{4}(2)\frac{1}{2}\sin(2x) = -\frac{\sin(2x)}{4}. Combining all parts, we get: f(x)=(x23x)12sin(2x)+14(2x3)cos(2x)+sin(2x)4+Cf(x) = (x^2 - 3x)\frac{1}{2}\sin(2x) + \frac{1}{4}(2x - 3)\cos(2x) + \frac{\sin(2x)}{4} + C, where CC is the constant of integration. Using the initial condition f(0)=5f(0) = 5, we find CC: f(0)=(0230)12sin(20)+14(203)cos(20)+sin(20)4+C=5f(0) = (0^2 - 3\cdot 0)\frac{1}{2}\sin(2\cdot 0) + \frac{1}{4}(2\cdot 0 - 3)\cos(2\cdot 0) + \frac{\sin(2\cdot 0)}{4} + C = 5. Since sin(0)=0\sin(0) = 0 and cos(0)=1\cos(0) = 1, this simplifies to: 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx00. Now we have the function 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx11. We can now evaluate f(6)f(6) using a calculator: 14cos(2x)(2)dx-\frac{1}{4}\cos(2x)(2)\,dx33. After calculating the above expression using a calculator, we find the value of f(6)f(6) to the nearest thousandth.

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