Differentiation Rules - Products~

sp09ta
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1. Suppose the curve f(x)=(x^4)+a(x^3)+b(x^2)+cx+d has tangent line when x=0 with equation y=2x+1, and a tangent line when x=1 with equation y=2-3x. Find a,b,c,d
2. d/dx f(x)g(x)=g*f`+f*g`
3. f`(x)=4x^3+3ax^2+2bx+c

x=0 y=2x+1 y=1 (0,1) is a point on f(x)
x=1 y=2x-3 y=-1 (1,-1) is a point on f(x)

This question really has me stumped... I just need a hint to get the ball rolling...
 
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sp09ta said:
1. Suppose the curve f(x)=(x^4)+a(x^3)+b(x^2)+cx+d has tangent line when x=0 with equation y=2x+1, and a tangent line when x=1 with equation y=2-3x. Find a,b,c,d



2. d/dx f(x)g(x)=g*f`+f*g`



3. f`(x)=4x^3+3ax^2+2bx+c

x=0 y=2x+1 y=1 (0,1) is a point on f(x)

Okay, so f(0)= d= 1 and f'(0)= c= the slope of y= 2x+1.

x=1 y=2x-3 y=-1 (1,-1) is a point on f(x)
so f(1)= ... and f'(1)= ...

This question really has me stumped... I just need a hint to get the ball rolling...
 
Ok, then f(1)=a+b+4=-1 and f'(1)=3a+2b+6=-3

so, a+b+5=0 and 3a+2b+9=0

then I isolate b in the first equation and sub it into the second to get:
3a+2(-a-5)+9=0
a=1

Then use a to find the value of b

So my values are: a=1, b=-6, c=2, d=1
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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