Linear algebra - find polynomial basis for vector space with conditions?

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Eagerissac

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Let S contain vectors of the form f(x) = a + bx + cx^2 + dx^3: f(-2) = 0 and f'(4) = 0

Find the basis for S.

What I attempted to do was plug in f(-2) = 0 first in the equation, I got:

a + b(-2) + c(-2)^2 + d(-2)^3. I assumed a would automatically = 0 because of the condition of f(-2) = 0

This leaves bx + cx^2 + dx^3

I then try to plug in f'(4) = 0 which makes b = 0 cause of the derivative part. This leaves 2c(4) + 3d(4)^2 = 8c + 48d.

This makes a = 0. b = 0. c = 8. and d = 48.

And I think my final answer would be 0 + 0x + 8x^2 + 48x^3 but I'm told it's wrong. Can someone help explain what I'm doing wrong?
 
a-2b+4c-8d=0
b+8c+48d=0
4 variables and 2 equations. You will need a pair of independent functions.
You may want to try using c=0 and d=1 for one and c=1 and d=0 for the other
 
a is 0 because the function is 0? What?

Why did you bother to write a + b(-2) + c(-2)^2 + d(-2)^3 ?????

f(-2) = a + b(-2) + c(-2)^2 + d(-2)^3 AND f(-2) = 0. So a + b(-2) + c(-2)^2 + d(-2)^3 = 0. That is a-2b+4c-8d=0.

Now find f'(x) so then you can find f'(4) OR if you prefer you can use the limit definition at a point to find f'(4) without finding f'(x). Then set this equation to 0.

No you have 2 equations. What do you do now?
 
Eagerissac said:
What I attempted to do was plug in f(-2) = 0 first in the equation
Click to expand...
What does that mean?

f(-2) is a constant!! How can it be possibly be bx + cx^2 + dx^3??

You do realize that a prerequisite for linear algebra is algebra, correct?

 
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