Find values of a,b,c and d such that g(x) = ax^3+bx^2+cx+d has a local maximum of (2,4) and a local minimum at (0,0).
How do I do this?
$\displaystyle f'(x)=3ax^2+2bx+c$...we know that $\displaystyle f'(0)=0$...therfore $\displaystyle 0=3a0^2+2b0+c,c=0$...now you also know that $\displaystyle f'(2)=0$...so $\displaystyle 0=3ax^2+2bx$...now that you get the jist..set up your other equations and solve for the coefficents
$\displaystyle f(x)=ax^3+bx^2+cx+d \mbox{ and } f'(x)=3ax^2+2bx+c$
With the above info we get
using the point (0,0)
$\displaystyle f(0)=0=a(0)^3+b(0)^2+c(0)+d \iff d=0$
$\displaystyle f'(0)=0=3a(0)^2+2b(0)+c \iff c=0$
since we know that c=d=0 we get
$\displaystyle f(x)=ax^3+bx^2 \mbox{ and } f'(x)=3ax^2+2bx$
using the point (2,4)
$\displaystyle f(2)=4=a(2)^3+b(2)^2 \iff 4=8a+4b$
$\displaystyle f'(2)=0=3a(2)^2+2b(2) \iff 0=12a+4b \iff b=-3a$
solving this system by substition (or elimination)
$\displaystyle 4 =8a+4(-3a) \iff 4 =-4a \iff a=-1$
so then b=3
Finally we get
$\displaystyle f(x)=-x^3+3x^2$