Pure Maths - functions question?

Question

Could you please explain:
A curve has equation Y=f(x). It is given that f'(x) = x^-3/2 + 1 and that f(4) = 5. Find f(x)

Thank you

Answer 1

Integration of the derivative function is equals to the function.

Given : f ' (x) = x^{(- 3/2)} + 1 and f(4) = 5.

ʃ f ' (x) dx = f (x).

ʃ f ' (x) dx = ʃ [x^{(- 3/2)} + 1] dx

= ʃ x^{(- 3/2)} dx + ʃ 1 dx

Apply formula : ʃ xⁿ dx = x^{(n + 1)}/(n + 1) + C

= x^{(- 3/2 + 1)} / (- 3/2 + 1) + x + C

= - 2x^{(- 1/2)} + x + C

= f (x).

So, f (x) = - 2x^{(- 1/2)} + x + C.

f (4) = - 2*4^{(- 1/2)} + 4 + C = 5     (From the given data)

- 2*2^{(- 2*1/2)} + C = 5 - 4 = 1

- 2*2^{(- 1)} + C = 1

- 2/2 + C = 1

- 1 + C = 1

⇒ C = 1 + 1 = 2.

f (x) = - 2x^{(- 1/2)} + x + 2.

Therefore, the function f (x) = - 2x^{(- 1/2)} + x + 2.