# Differentiation help

• Apr 22nd 2007, 12:11 PM
johnnyboy
Differentiation help
I have a problem with this exercise:

"The curve with equation: y = ax^3 + bx^2 + cx + d has a local maximum point where x=-3 and a local minimum point where x=-1. If the curve passes through the points (0;4) and (1;20), find the equation for y in terms of x"

I get that "d = 4"; "a + b + c=16"; "27a - 6b + c" and "3a - 2b + c" both are equal to 0; and that "2b - 18a is less than zero" while "2b - 6a is greater than zero"

But I'm stuck, and can't seem to find the values for a;b and c.

Thanks
• Apr 22nd 2007, 12:55 PM
Jhevon
Quote:

Originally Posted by johnnyboy
I have a problem with this exercise:

"The curve with equation: y = ax^3 + bx^2 + cx + d has a local maximum point where x=-3 and a local minimum point where x=-1. If the curve passes through the points (0;4) and (1;20), find the equation for y in terms of x"

I get that "d = 4"; "a + b + c=16"; "27a - 6b + c" and "3a - 2b + c" both are equal to 0; and that "2b - 18a is less than zero" while "2b - 6a is greater than zero"

But I'm stuck, and can't seem to find the values for a;b and c.

Thanks

we know that y passes through (0,4) and (1,20)
y = ax^3 + bx^2 + cx + d

=> 4 = a(0)^3 + b(0)^2 + c(0) + d
=> d = 4

=> 20 = a(1)^3 + b(1)^2 + c(1) + d
=> 20 = a + b + c + 4
=> 16 = a + b + c
that is, a + b + c = 16
call this equation (1)

so a + b + c = 16 .....................(1)

we also know that the derivatve gives the slope at any value of x right? so let's start there

y = ax^3 + bx^2 + cx + d
=> y' = 3ax^2 + 2bx + c
we have extrema occuring at the points where y' = 0

set y' = 0
=> 3ax^2 + 2bx + c = 0

we know that x = -1 and x = -3, so we have the simultaneous equations:

3a(-1)^2 + 2b(-1) + c = 0 ....................(2)
3a(-3)^2 + 2b(-3) + c = 0 ....................(3)

3a - 2b + c = 0 ....................(2)
27a - 6b + c = 0 ..................(3)

so we know d, to find a,b, and c we must solve the system:

a + b + c = 16 .....................(1)
3a - 2b + c = 0 ....................(2)
27a - 6b + c = 0 ..................(3)

=> 2a - 3b = -16 ...................(4) = (2) - (1)
=> 24a - 4b = 0 .....................(5) = (3) - (2)

=> 24a - 36b = -192 ...............(4)*12
=> 24a - 4b = 0 .....................(5)

=> 32b = 192 .........................(5) - (4)
=> b = 6

but 2a - 3b = -16
=> 2a - 18 = -16
=> 2a = 2
=> a = 1

but a + b + c = 16
=> 1 + 6 + c = 16
=> c = 16 - 7 = 9

so we have a = 1, b = 6, c = 9, d = 4

so the curve is y = x^3 + 6x^2 + 9x + 4
• Apr 22nd 2007, 12:58 PM
johnnyboy
thanks a lot. It seems so obvious now