Fun with Cubic's

[jado818]

Ive been trying to figure this problem out for about a week and I can't seem to find the trick to it....

Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 3 at -2 and a local minimum value of 0 at 1.


Now I can figure out from the given information four equations because it gives me two points to work with Max at (-2,3) and Min at (1,0)

first of all

dy/dx = 3ax^2 + 2bx + c

the four equations that I have are

f(-2) = -8a + 4b -2c +d
f(1 ) = a + b + c + d

dy/dx f(-2) = 12a - 4b + c
dy/dx f(1) = 3a + 2b + c

Now if i could just figure out how to solve for what a,b,c, and d equal I could write the equation.

It sounds simple but i can't figure it out

any help would be appreciated

Jack

[HallsofIvy]

Ive been trying to figure this problem out for about a week and I can't seem to find the trick to it....

Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 3 at -2 and a local minimum value of 0 at 1.


Now I can figure out from the given information four equations because it gives me two points to work with Max at (-2,3) and Min at (1,0)

first of all

dy/dx = 3ax^2 + 2bx + c

the four equations that I have are

f(-2) = -8a + 4b -2c +d
f(1 ) = a + b + c + d

dy/dx f(-2) = 12a - 4b + c
dy/dx f(1) = 3a + 2b + c

Now if i could just figure out how to solve for what a,b,c, and d equal I could write the equation.

It sounds simple but i can't figure it out

any help would be appreciated

Jack
You didn't finish writing what you know:
f(-2) = -8a + 4b -2c +d= 3 since maximum value is 3
f(1 ) = a + b + c + d= 0 since the minimum value is 0

dy/dx f(-2) = 12a - 4b + c= 0 since -2 is a critical point
dy/dx f(1) = 3a + 2b + c= 0 since 1 is a criticl point.
Subtracting a+ b+ c+ d= 0 from -8a+ 4b- 2c+ 2= 3 gives
-9a+ 3b- 3c= 3 or -3a+ b- c= 1, eliminating d.

Adding 3a+ 2b+ c= 0 to that gives 3b= 1 which happens to eliminate both a and c in one fell swoop! b= 1/3. Putting that back into 3a+ 2b+ c= 0 gives
3a+ c= -2/3. Putting b= 1/3 into 12a- 4b+ c= 0 gives 12a+ c= 4/3. Now subtract 3a+ c= -2/3 from that: 9a= 6/3= 2 so a= 2/9. Putting that into 3a+ c= -2/3 gives 2/3+ c= -2/3 so c= -4/3. Finally, putting all of those into a+ b+ c+ d= 0, 2/9+ 1/3- 4/3+ d= 0 so d= 1- 2/9= 7/9. The cubic function is
y= (2/9)x^3+ (1/3)x^2- (4/3)x+ 7/9.

[jado818]

Thanks for the help