Question

f(x) = ax^3+bx^2+cx+d

Answer 1

find a,b,c,d so that the function has a local maximum at f(-1) = 2 and has a local minimum at f(1) = -1

Answer 2

for a local maximum, f'(x)=0 and f''(x)<0
for a local minimum, f'(x)=0 and f''(x)>0

so you need to calculate f'(x) and f''(x):\[\begin{align}
fx)&=ax^3+bx^2+cx+d\\
f'(x)&=3ax^2+2bx+c\\
f''(x)&=6ax+2b
\end{align}\]
you are told local maximum is at x=-1, so we know f'(-1)=0 and f''(-1)<0 giving:\[\begin{align}
f'(-1)&=3a*(-1)^2+2b*(-1)+c&=0\\
&\therefore 3a-2b+c&=0\tag{1}\\
f''(-1)&=6a*(-1)+2b&<0\\
&\therefore -6a+2b&<0\tag{2}
\end{align}\]similarly, you are told local minimum is at x=1, so we know f'(1)=0 and f''(1)>0 giving:\[\begin{align}
f'(1)&=3a*(1)^2+2b*(1)+c&=0\\
&\therefore 3a+2b+c&=0\tag{3}\\
f''(1)&=6a*(1)+2b&>0\\
&\therefore 6a+2b&>0\tag{4}
\end{align}\]finally, you are also told that f(-1)=2 and f(1)=-1 giving:\[\begin{align}
f(-1)&=a*(-1)^3+b*(-1)^2+c*(-1)+d&=2\\
&\therefore -a+b-c+d&=2\tag{5}\\
f(1)&=a*(1)^3+b*(1)^2+c*(1)+d&=-1\\
&\therefore a+b+c+d&=-1\tag{5}
\end{align}\]
you should be able to use these 5 equations to find a, b, c and d.