Write the general cubic function:
#f(x) = ax^3+bx^2+cx+d#
Its first and second derivative are:
#f'(x) =3ax^2+2bx +c#
#f''(x) = 6ax +2b#
If #f(x)# has an inflection point in #(3,2)# this means that:
#f''(3) = 0 => color(blue)(18a + 2b = 0)#
and
#f(3) = 2 => color(blue)(27a +9b +3c + d = 2)#
Then we know that the point #(5,10)# is a maximum, so:
#f'(5) = 0 => color(blue)(75a+10b+c =0)#
#f(5) = 10 => color(blue)(125a+25b+5c+d= 10)#
We have now four linear equations in four unknowns: the solution of the system gives us the values of the coefficients of the cubic:
#{(18a + 2b = 0),(27a +9b +3c + d = 2),(75a+10b+c =0),(125a+25b+5c+d= 10):}#
Subtract the second equation from the fourth:
#{(18a + 2b = 0),(98a +16b +2c =8),(75a+10b+c =0),(125a+25b+5c+d= 10):}#
Multiply the third equation by #2#:
#{(18a + 2b = 0),(98a +16b +2c =8),(150a+20b+2c =0),(125a+25b+5c+d= 10):}#
Subtract the second from the third:
#{(18a + 2b = 0),(52a +4b =- 8),(150a+20b+2c =0),(125a+25b+5c+d= 10):}#
Multiply the first equation by #2#:
#{(36a + 4b = 0),(52a +4b =- 8),(150a+20b+2c =0),(125a+25b+5c+d= 10):}#
Subtract the first from the second:
#{(16a = -8),(52a +4b =- 8),(150a+20b+2c =0),(125a+25b+5c+d= 10):}#
and we have:
#{(a = -1/2),(b=9/2),(c=-15/2),(d=-5/2):}#
So the function is:
#f(x) = -1/2x^3+9/2x^2-15/2x-5/2#
graph{-1/2x^3+9/2x^2-15/2x-5/2 [-29.24, 34.84, -14.1, 17.94]}
