Find a cubic function
y = ax^3 + bx^2 + cx + d
whose graph has horizontal tangents at the points
(−2, 10) and (2, 4)
Not sure what to do, first time I have ever seen a question like this
y = ax^3 + bx^2 + cx + d
whose graph has horizontal tangents at the points
(−2, 10) and (2, 4)
Not sure what to do, first time I have ever seen a question like this
-
dy/dx=3ax²+2bx+c
And dy/dx=0 when x=-2 and 2
You need to form two equations by substituting the x values into the original equation.
These are 10=-8a+4b-2c+d
and 4=8a+4b+2c+d
Then take dy/dx and get 3ax²+2bx+c
This =0 when x=-2 and 2
so 12a-4b+c=0 and 12a+4b+c=0
You can see from this that b=0, and the equation reduces to 12a+c=0.
You then have to use this, and the equations you have to work the others out.
There IS enough information, despite Ted's doubts.
Your final answer should be y=(3/16)x³-(9/4)x+7
Interesting question, more involved than it seemed!
And dy/dx=0 when x=-2 and 2
You need to form two equations by substituting the x values into the original equation.
These are 10=-8a+4b-2c+d
and 4=8a+4b+2c+d
Then take dy/dx and get 3ax²+2bx+c
This =0 when x=-2 and 2
so 12a-4b+c=0 and 12a+4b+c=0
You can see from this that b=0, and the equation reduces to 12a+c=0.
You then have to use this, and the equations you have to work the others out.
There IS enough information, despite Ted's doubts.
Your final answer should be y=(3/16)x³-(9/4)x+7
Interesting question, more involved than it seemed!
-
not sure it you have enough information
dy / dx = 0 at x = - 2 and 2...thus dy /dx = µ ( x² - 4 )--> µ = 3 a and µ = - c / 4 & b = 0
also 10 = - 8 a + 4 b - 2c + d and 4 = 8a + 4b + 2c + d---> 14 = 8 b + 2d &
6 = - 16 a - 4 c ...thus d = 7 and 6 = -16 [ µ / 3 ] - 4 [ - 4 µ ]
was enough ..solve for µ , then a and c
dy / dx = 0 at x = - 2 and 2...thus dy /dx = µ ( x² - 4 )--> µ = 3 a and µ = - c / 4 & b = 0
also 10 = - 8 a + 4 b - 2c + d and 4 = 8a + 4b + 2c + d---> 14 = 8 b + 2d &
6 = - 16 a - 4 c ...thus d = 7 and 6 = -16 [ µ / 3 ] - 4 [ - 4 µ ]
was enough ..solve for µ , then a and c