derivatives?

Question

if f"(x) = 3/sqrt(x) and f'(4) =7 then f(x) equals

Answer 1

f"(x)=3x^(-1/2) f'(x)= (1/ 1/2)3x^(1/2) +c =6x^(1/2) +c

f'(4)=7 6(4) +c=7 c= -5 so, f'(x)=6x^(1/2)-5

f(x)=6(1/ 3/2)x^(3/2) -5x+c = 4x^(3/2)-5x+c

Answer 2

Integral 3/sqrt x dx = f(x) =

3(Integral dx/sqrt x) =

3(Integral x^-1/2 dx) =

3(2x^1/2) + C

6sqrt x + C = f(x)

f'(4) CANNOT equal 7, so I assume you meant f(4) = 7

6(sqrt 4) + C = 7

6(2) + C = 7

12 + C = 7

C = -5

f(x) = 6sqrt x - 5

Answer 3

if f"(x) = 3/sqrt(x) and f'(4) =7 then f(x) equals

your question is not derivatives

Answer 4

f'(x)=integral of 3/sqrt(x)

=3 * 2 sqrt(x)+C
7=6sqrt(4)+C
C=7-12=-5

f'(x)=6sqrt(x)-5
take integral again

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

Answer 5

f "(x) = 3 x^(-1/2)
f `(x) = 6 x^(1/2) + C
f `(4) = 12 + C
7 = 12 + C
C = - 5
f `(x) = 6 x^(1/2) - 5
f (x) = 4 x^(3/2) - 5 x + C

Answer 6

integrating f''(x) we get

f'(x)=int(3/sqrt(x))
=int(3x^-1/2)
=6x^1/2+C
=6sqrt(x)+C

Substituting f'(4)=7 :

7=6sqrt(4)+C
7=12+C
C=-5

f'(x) = 6sqrt(x)-5

Integrating again:

f(x) = 4x^3/2 - 5x + C