Question 1106177
<br>
We know that...<br>
(a) f(3) = 1:
{{{(1/3)(3^3)+a(3^2)+b(3) = 1}}}
{{{9+9a+3b = 1}}}
{{{9a+3b = -8}}}  (1)<br>
and<br>
(b) f'(3) = 0:
f'(x) = x^2+2ax+b
{{{9+6a+b = 0}}}
{{{6a+b = -9}}}  (2)<br>
Solve the pair of equations (1) and (2).<br>
The numbers turn out to be ugly fractions, so solve for both a and b using elimination.<br>
solve for a (eliminate b):
{{{9a+3b = -8}}}
{{{18a+3b = -27}}}
{{{9a = -19}}}
{{{a = -19/9}}}
solve for b (eliminate a):
{{{18a+6b = -16}}}
{{{18a+3b = -27}}}
{{{3b = 11}}}
{{{b = 11/3}}}<br>
The required function is {{{f(x) = (1/3)x^3-(19/9)x^2+(11/3)x}}}<br>
The graph shows a relative minimum at (3,1):<br>
{{{graph(400,400,-2,8,-10,10,(1/3)x^3-(19/9)x^2+(11/3)x)}}}