Find f by solving the initial value (2 questions)

Calculus
Tutor: None Selected Time limit: 1 Day

Jul 23rd, 2015

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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

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 23rd, 2015

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