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{:[f^(')(x)=8x^(3)-12 x" and "f(1)=1],[f(-2)=]:}

f(x)=8x312x and f(1)=1f(2)= \begin{array}{l}f^{\prime}(x)=8 x^{3}-12 x \text { and } f(1)=1 \\ f(-2)=\end{array}

Full solution

Q. f(x)=8x312x and f(1)=1f(2)= \begin{array}{l}f^{\prime}(x)=8 x^{3}-12 x \text { and } f(1)=1 \\ f(-2)=\end{array}
  1. Calculate f(1)f(1): Calculate f(1)f(1) directly from the given information.\newlinef(1)=1f(1) = 1
  2. Integrate f(x)f'(x): To find f(2)f(-2), integrate f(x)f'(x) to get f(x)f(x). The integral of f(x)=8x312xf'(x) = 8x^3 - 12x is:\newline(8x312x)dx=2x46x2+C\int(8x^3 - 12x) dx = 2x^4 - 6x^2 + C, where CC is the constant of integration.
  3. Find Constant CC: Use the given f(1)=1f(1) = 1 to find CC.
    1=2(1)46(1)2+C1 = 2(1)^4 - 6(1)^2 + C
    1=26+C1 = 2 - 6 + C
    C=5C = 5
  4. Substitute x=2x = -2: Now substitute x=2x = -2 into f(x)=2x46x2+5f(x) = 2x^4 - 6x^2 + 5 to find f(2)f(-2).
    f(2)=2(2)46(2)2+5f(-2) = 2(-2)^4 - 6(-2)^2 + 5
    f(2)=2(16)6(4)+5f(-2) = 2(16) - 6(4) + 5
    f(2)=3224+5f(-2) = 32 - 24 + 5
    f(2)=13f(-2) = 13

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