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answers:
Ian H say: Let T = f(x) + f'(x)+ f"(x)
If f(x) = x^3 + ax^2 + bx + c
f '(x) = ………..3x^2 + 2ax + b
f"(x) = ……………………….6x + 2a
T = x^3 +(a + 3)x^2 + (2a + b + 6)x + 2a + b + c, compare with
T = x3 …+………3x^2..+………………..3x …-……….1
a = 0
b = -3
0 – 3 + c = -1, so c = 2
f(x) = x^3 - 3x + 2
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mbbs say: Well, evaluating a function means plugging whatever they gave me in for the argument in the formula. This means that I have to plug this character "§" in for every instance of x. Here goes:
f (§) = (§)2 + 2(§) – 1
= §2 + 2§ – 1
This is definitely weird-looking, but it follows all the rules they gave me, so:
f (§) = §2 + 2§ – 1
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The above example is kinda pointless, I'll grant you, but it clearly illustrates how the notation works. You plug the given value in for the given variable, and chug your way to the answer. Hence, these exercises are often referred to as "plug-n-chug". Your best bet is to try not to over-think them.
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Joseph say: find it yourself
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CogitoErgoCogitoSum say: Assuming f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
f''(x) = 6ax + 2b
Add it all up.
ax^3 + (3a+b)x^2 + (6a + 2b + c)x + (2b+c+d)
Set it equal to x^3 + 3x^2 + 3x - 1
ax^3 + (3a+b)x^2 + (6a + 2b + c)x + (2b+c+d) = x^3 + 3x^2 + 3x - 1
And compare like-terms
a = 1
3a+b = 3
6a + 2b + c = 3
2b+c+d = -1
Given that a=1 it follows that b=0, then that c = -3, and lastly that d=2.
So,
f(x) = x^3 - 3x + 2
This is the non-homogeneous solution. Still need to find the homogeneous solutions.
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cidyah say: y+y'+y'' = x^3+3x^2+3x-1
Complementary solution: