Differentiation Rules - Products~

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SUMMARY

The discussion focuses on finding the coefficients a, b, c, and d for the polynomial function f(x) = x^4 + a(x^3) + b(x^2) + cx + d, given specific tangent lines at x=0 and x=1. The tangent line at x=0 is y=2x+1, indicating that f(0)=1 and f'(0)=2. The tangent line at x=1 is y=2-3x, leading to f(1)=-1 and f'(1)=-3. Through solving the resulting equations, the values are determined as a=1, b=-6, c=2, and d=1.

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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
 

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