calculus prob?

Question

Find f(x) if f"(x) = 6x −6/x4 , and f'(1) = 8, and f(1) = 5

Answer 1

f´(x) = 3x^2+2 x^-3+c
8= 3+2+c so c= 3 and f´(x)=3x^2+2x^-3+3
f(x) =x^3-x^-2+3x +c
5=3+c so c =2
f(x)= x^3-x^-2+3x+2

Answer 2

f'(x) = 8 + ∫6x −6/x4 dx, from 1 to x = 8 + 3(x^2-1) + 2(1/x^3 - 1)
=> f'(x) = 3x^2 + 2/x^3 + 3

f(x) = 5 + ∫3x^2 + 2/x^3 + 3 dx, from 1 to x = 5 + (x^3 -1) - (1/x^2 -1) + 3(x -1)
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Ideas to save time: Use definite integral to include boundaries.

Answer 3

f''(x) = 6x - 6/x^4

integrating once

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

f'(x) = 3x^2 + 2/x^3 + c

f'(1) = 3 + 2 + c => 5 + c

so 5 + c = 8

c = 3

f'(x) = 3x^2 + 2/x^3 + 3

again integrating

f(x) = 3x^3/3 + 2(x^-2)/-2 + 3x + c1

f(x) = x^3 - 1/x^2 + 3x + c1

f(1) = 1 - 1 + 3 + c1

c1 + 3 = 5

c1 = 2

f(x) = x^3 - 1/x^2 + 3x + 2

f(x) = [x^5 - 1 + 3x^3 + 2x^2]/x^2

f(x) = (x^5 + 3x^3 + 2x^2 - 1)/x^2

Answer 4

start with f''
antiderivitive we got f' = 3x^2 + 2x^-3 + c f'(1)= 8

subtitude in solve for c and you get c = 3

for f'(x) = 3x^2 + 2x^-3 + 3

antiderivitive again f(x)= x^3 - x^-2 + 3x + c f(1) = 5
subtitude in you got c = 2

so f(x) = x^3 - 1/x^2 +3x+2

Answer 5

dy/dx means "by-made from y (that's possibly a function of x) with appreciate to x." given which you do not actually have a y right here, that's complicated to do. yet, i'm going to tell you dx/dt: dx/dt = 4 d/dt (sin t) - 2 d/dt (tan 2t) = 4 cos t - 2 sec^2 (2t) * d/dt (2t) = 4 cos t - 4 sec^2 (2t)

Answer 6

Brings back bad memories.

If I had payed attention back then (which was only last year) I can help you out. Sorry.