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?

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- Apr 10th 2008, 07:09 PMtheowneFind function with these minimums and maximums
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? - Apr 10th 2008, 07:11 PMMathstud28
- Apr 10th 2008, 07:13 PMtheowne
Yes, that is the question...and what to do from there to find the original equation?

- Apr 10th 2008, 07:15 PMMathstud28Uh...
- Apr 10th 2008, 07:46 PMtheowne
Set what up? What four equations? The only equations I can see involved are the function, derivative and second derivative. I am attempting to learn this material, it'd be more helpful to see an example to learn from.

- Apr 10th 2008, 08:05 PMMathstud28Ok
$\displaystyle f'(x)=3ax^2+2bx+c$...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

- Apr 10th 2008, 09:35 PMTheEmptySet
$\displaystyle f(x)=ax^3+bx^2+cx+d \mbox{ and } f'(x)=3ax^2+2bx+c$

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

$\displaystyle f(x)=-x^3+3x^2$