Please help with math

Question

1) Find f. f '''(x) = cos(x),    f(0) = 4,    f '(0) = 8,    f''(0) = 7

2) A particle is moving with the given data. Find the position of the particle, s(t).

a(t) = t^2 − 7t + 4, s(0) = 0, s(1) = 20

3) Find f. f ''(x) = 10 + 6x + 24x^2, f(0) = 5, f (1) = 17

Any help is appreciated

Answer 1

To go from a derivative back to a function, you must consider the derivative rules backwards. So if f '''(x) = cos(x) you need to consider what function could the second derivative be that will result in cos(x) if I take its derivative. That would be sin(x) so f ''(x) = sin(x) + C (because the derivative of any constant is zero). To find the value of "C", use the condition given that f ''(0) = 7. so 7 = sin(0) + C or C = 7.

So f ''(x) = sin(x) + 7

Now, what would f '(x) need to be that will result in sin(x) + 7 when you take its derivative? f '(x) must be -cos(x) + 7x + C. To find C, use the condition given that f '(0) = 8 so: 8 = -cos(0) + 7(0) + C or C = 9.

So f '(x) = -cos(x) + 7x + 9

Now, what would f(x) need to be that will result in -cos(x) + 7x + 9 when you take its derivative? f(x) must be -sin(x) + 7/2x² + 9x + C and to find C we use the condition given that f (0) = 4. So 4 = -sin(0) + 7/2(0)² + 9(0) + C meaning C = 4

So f(x) = -sin(x) + 7/2x² + 9x + 4

Use the same approach for both 2) and 3).