If f'(x) = 8x^3 + 3x^2 + 2 , find f(x) if f(0) = -5.
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f'(x) = 8x^3 + 3x^2 + 2
We know that f(x) = integral of f'(x).
==> f(x) = intg (8x^3 + 3x^2 + 2) dx
= intg 8x^3 dx + intg 3x^2 dx + intg 2 dx
= 8x^4/4 + 3x^3/3 + 2x + C
==> f(x) = 2x^4 + x^3 + 2x + C
But we are given that f(0) = -5
==> f(0) = 0+ C = -5
==> C = -5
==> f(x) = 2x^4 + x^3 + 2x -5
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We are given that f'(x) = 8x^3 + 3x^2 + 2. To find f(x) we need to find the integral of f'(x).
Int[f'(x)] = Int [ 8x^3 + 3x^2 + 2]
=> 8x^4 / 4 + 3x^3 / 3 + 2x + C
=> 2x^4 + x^3 + 2x + C
Also, f(0) = -5
2*0^4 + 0^3 + 2*0 + C = -5
=> C = -5
Therefore f(x) = 2x^4 + x^3 + 2x - 5
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