Find f by solving the initial value (2 questions)
Calculus

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11) f '(x) = 6x^2  8x , f(1) = 6
We just need to integrate by using these formulas:
integral(x^n dx) = x^n+1/n+1 + C ; where: n is the exponent of the x and C is an arbitrary constant of integration.
integral(K dx) = K*x + C ; k is any real number.
f(x) = integral( (6x^2  8x) dx) = integral( (6x^2  8x^1) dx) = 6x^2+1/2+1  8x^1+1/1+1 + C
f(x) = 6x^3/3  8x^2/2 + C = 2x^3  4x^2 + C
Then we find the value of C.
f(1) = 6 > 2(1^3)  4(1^2) + C = 6 > 2(1)  4(1) + C = 6 > 2  4 + C = 6
2 + C = 6 >  2 + 2 + C = 6 + 2 > C = 8
Final answer: f(x) = 2x^3  4x^2 + 8
12) f '(x) = 6e^x  6x
Let's remember that the antiderivative or integral of the exponential function:
integral( e^x dx ) = e^x + C
So then we would have:
f(x) = integral( (6e^x  6x) dx ) = integral( (6e^x  6x^1)dx ) = 6e^x  6x^1+1/1+1 + C = 6e^x  6x^2/2 + C
f(x) = 6e^x  3x^2 + C
f(0) = 8 > 6e^0  3(0^2) + C = 8 > 6(1)  3(0) + C = 8 > 6  0 + C = 8
6 + C = 8 > 6  6 + C = 8  6 > C = 2
Final answer: f(x) = 6e^x  3x^2 + 2
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