Find a cubic function: max of 3 at x=-2, min of 0 at x=1

[dangerous_dave]

I have another question
Find a cubic function that has a local maximum of 3 at x=-2 and a local minimum of 0 at x=1.

This one I don't even know how to start...

Thanks

 

[soroban]

Hello, dangerous_dave!

Find a cubic function that has a local maximum of 3 at x = -2
and a local minimum of 0 at x = 1.

$\displaystyle \text{The general cubic function is: }\;f(x) \;=\;ax^3 + bx^2 + cx + d$
. . $\displaystyle \text{and we must determine }a,b,c,d.$

We know the point (-2,3) is on the cubic: .$\displaystyle f(-2) \:=\:3$
. . So we have: . $\displaystyle -8a + 4b - 2c + d \:=\:3$ . [1]

We know the point (1,0) is on the cubic: .$\displaystyle f(1) \:=\:0$
. . So we have: . $\displaystyle a + b + c + d \:=\:0$ . [2]


The derivative is: . $\displaystyle f'(x) \;=\;3ax^2 + 2bx + c$
The cubic has critical values at $\displaystyle x \:=\:-2,\:1$
. . Hence, the derivative equals zero at those values.

. . $\displaystyle f'(\text{-}2) = 0:\;\;12a - 4b + c \:=\:0$ . [3]
. . .$\displaystyle f(1) \:=\:0:\;\;\;3a + 2b + c \:=\:0$ . [4]


We have a system of four equations.
. . $\displaystyle \text{Now solve for }a,b,c,d.$

 

[dangerous_dave]

Hello, soroban!

Thankyou for that. It helps a lot .