Let f′′′(t)=−5t+5√t

Question

(a) Find the most general formula for f′′(t). If an arbitrary constant must be used here, use an upper-case "C".

 

(b) Based on your answer to (a), find the most general formula  for f′(t). If another new arbitrary constant must be used here, use an upper-case "D".

 

(c) Based on your answer to (b), find the most general formula for f(t). If another new arbitrary constant must be used here, use an upper-case "E".

 

Answer 1

f'''(t) = -5t + 5* t^(1/2)

f''(t) = (-5/2)t^2 + (2/3)(5) t^(3/2)

= (-5/2) t^2 + (10/3) t^(3/2) + C

f'(t) = (-5/6)t^3 + (2/5)(10/3) t^(5/2) + Cx + D

= (-5/6)t^3 + (4/3) t^(5/2) + Cx + D

f(t) = (-5/24)t^4 + (2/7)(4/3) t^(7/2) + (1/2)Cx^2 + Dx+E

= (-5/24)t^4 + (8/21) t^7/2 + (1/2)C x^2 + Dx + E

check by differentiation:

y = (-5/24)t^4 + (8/21) t^7/2 + (1/2)C x^2 + Dx + E

y' = (-5/6) t^3 + (4/3) t^(5/2) + Cx + D

y'' = (-5/2) t^2 + (10/3) t^(3/2) + C

y''' = -5 t + 5 t^(1/2)