please show work thank you Find the function f fx 4x 3x2 f

please show work thank you

Find the function f. f\"(x) = 4x - 3/x2, f(1) = 1/3, f\'(1) = 5

Solution

f\"(x)=4x-3/x²

So, f\'(x)=Integration of f\'\'(x)dx

=integration of (4x-3/x²)dx

=2x²+3/x +c

Given, f\'(1)=5

So, 5=2*(1)²+3/1 +c

or, 5=2+3+c

or, c=0

So, f\'(x)=2x²+3/x

So, f(x)=integration of f\'(x)dx

=integration of (2x²+3/x)dx

=2*(x³/3) +3lnx +D

=(2/3)x³ +3lnx +D

Given f(1)=1/3

So, 1/3=(2/3)(1)³+3ln(1)+D
or, 1/3=2/3+D

or, D=-1/3

So,

f(x)=(2/3)x³+3lnx-(1/3)