Find the values of an equation using derivatives?

Suppose the curve y = x^4 + ax^3 + bx^2 + cx + d has a tangent line when x = 0 with the equation y=2x+1 and a tangent line when x = 1 with the equation y=2-3x. Find the values of a, b, c, d.

I've already found c and d to be 2 and 1, respectively.

y = x^4 + ax3 + bx2 + cx + d

At x = 0, the tangent is y = 2x + 1, and y = 2(0) + 1 = 1

At x = 1, the tangent is y = -3x + 2, and y = -3(1) + 2 = -1

Write two independent equation given points (0,1) and (1,-1):

(0)^4 + a(0)3 + b(0)2 + c(0) + d = 1 ==> d = 1

(1)^4 + a(1)3 + b(1)2 + c(1) + 1 = -1 ==> a + b + c = -2

We can write two more independent equation using the derivative of the function:

dy/dx = 4x3 + 3ax2 + 2bx + c

At x = 0, dy/dx = 2:

2 = 4(0)3 + 3a(0)2 + 2b(0) + c

c = 2

At x = 1, dy/dx = -3:

-3 = 4(1)3 + 3a(1)2 + 2b(1) + c

3a + 2b + c = -7

Thus, the four equations are:

d = 1

a + b + c = -2

c = 2

3a + 2b + c = -7

Since c = 2, the second and fourth equations become:

a + b = -4

3a + 2b = -9

Using the method of elimination, multiply all terms in the first equation by -2 to make the b-terms equal and opposite:

-2a - 2b = 8

3a + 2b = -9

Now add terms vertically (the b-terms will cancel):

a = -1

Now solve for "b":

a + b = -4

-1 + b = -4

b = -3

Thus, the four coefficents are: a = -1, b = -3, c = 2, and d = 1, and the equation becomes:

y = x^4 - x3 - 3x2 + 2x + 1

#1

This a good exam question and there are several pitfalls you want to avoid in solving this type of problem.