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{:[f^(')(x)=6x^(5)+7" and "],[f(-2)=30.]:}

f(1)=

f(x)=6x5+7 and f(2)=30. \begin{array}{l}f^{\prime}(x)=6 x^{5}+7 \text { and } f(-2)=30 .\end{array} \newlinef(1)= f(1)=

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Q. f(x)=6x5+7 and f(2)=30. \begin{array}{l}f^{\prime}(x)=6 x^{5}+7 \text { and } f(-2)=30 .\end{array} \newlinef(1)= f(1)=
  1. Integrate f(x)f'(x): To find f(1)f(1), we need to integrate the derivative f(x)f'(x) to get the original function f(x)f(x). The derivative f(x)f'(x) is given as 6x5+76x^{5} + 7. Integrate f(x)f'(x) with respect to xx to find f(x)f(x). (6x5+7)dx=6x5dx+7dx=(66)x6+7x+C=x6+7x+C\int(6x^{5} + 7) dx = \int 6x^{5} dx + \int 7 dx = (\frac{6}{6})x^{6} + 7x + C = x^{6} + 7x + C, where f(1)f(1)00 is the constant of integration.
  2. Find Constant C: We know that f(2)=30f(-2) = 30. We can use this information to find the constant CC.\newlineSubstitute x=2x = -2 and f(x)=30f(x) = 30 into the equation f(x)=x6+7x+Cf(x) = x^{6} + 7x + C.\newline30=(2)6+7(2)+C30 = (-2)^{6} + 7(-2) + C\newline30=6414+C30 = 64 - 14 + C\newline30=50+C30 = 50 + C\newlineC=3050C = 30 - 50\newlineC=20C = -20
  3. Write Complete Function: Now that we have the constant CC, we can write the complete function f(x)f(x).f(x)=x6+7x20f(x) = x^{6} + 7x - 20
  4. Substitute x=1x = 1: To find f(1)f(1), substitute x=1x = 1 into the function f(x)f(x).
    f(1)=(1)6+7(1)20f(1) = (1)^{6} + 7(1) - 20
    f(1)=1+720f(1) = 1 + 7 - 20
    f(1)=820f(1) = 8 - 20
    f(1)=12f(1) = -12

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