Find f by solving the initial value (2 questions)

Calculus
Tutor: None Selected Time limit: 1 Day

Jul 23rd, 2015

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f(x)=6x^3/3-8x^2/2+c

f(x)=2x^3-4x^2+c

f(1)=6

so 

6=2*1-4*1+c

c=8

so f(x)=3x^3-4x^2+8


2)f(x)=6e^x-6x^2/2+c

f(x)=6e^x-3x^2+c

f(0)=8

8=6+c

c=2

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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