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 .