# Find function with these minimums and maximums

• Apr 10th 2008, 07:09 PM
theowne
Find 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 PM
Mathstud28
Quote:

Originally Posted by theowne
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?

You know four things $g(0)=0,g(2)=4,g'(2)=0,g'(0)=0$....haha and two more $f''(2)<0$...and $f''(0)>0$
• Apr 10th 2008, 07:13 PM
theowne
Yes, that is the question...and what to do from there to find the original equation?
• Apr 10th 2008, 07:15 PM
Mathstud28
Uh...
Quote:

Originally Posted by theowne
Yes, that is the question...and what to do from there to find the original equation?

...hmm four equations with four variables...hmm that sounds solvable...hmm...set it up =D haha...
• Apr 10th 2008, 07:46 PM
theowne
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 PM
Mathstud28
Ok
Quote:

Originally Posted by theowne
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.

$f'(x)=3ax^2+2bx+c$...we know that $f'(0)=0$...therfore $0=3a0^2+2b0+c,c=0$...now you also know that $f'(2)=0$...so $0=3ax^2+2bx$...now that you get the jist..set up your other equations and solve for the coefficents
• Apr 10th 2008, 09:35 PM
TheEmptySet
Quote:

Originally Posted by theowne
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?

$f(x)=ax^3+bx^2+cx+d \mbox{ and } f'(x)=3ax^2+2bx+c$

With the above info we get

using the point (0,0)

$f(0)=0=a(0)^3+b(0)^2+c(0)+d \iff d=0$

$f'(0)=0=3a(0)^2+2b(0)+c \iff c=0$

since we know that c=d=0 we get

$f(x)=ax^3+bx^2 \mbox{ and } f'(x)=3ax^2+2bx$

using the point (2,4)

$f(2)=4=a(2)^3+b(2)^2 \iff 4=8a+4b$

$f'(2)=0=3a(2)^2+2b(2) \iff 0=12a+4b \iff b=-3a$

solving this system by substition (or elimination)

$4 =8a+4(-3a) \iff 4 =-4a \iff a=-1$

so then b=3

Finally we get

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