Find T(1), T(x), T(x 2 ), and hence show that T(ax 2 +bx +c)=ax 2 - (a-2b+2c)x +
ID: 2943753 • Letter: F
Question
Find T(1), T(x), T(x2), and hence show that T(ax2 +bx +c)=ax2 - (a-2b+2c)x + 3c
where a,b,c are arbitrary real numbers.
Explanation / Answer
Since T is a linear transformation we can use the additivity and homogeneity properties. 1. The easiest one to get is T(x). Here, T(2x) = 4x --> 2T(x) = 4x --> T(x) = 2x 2. To get T(1) take your second and third equations. Multiply the first one by -3 and the second one by 2: -3 T(2x) = -12x --> T(-6x) = -12x 2 T(3x + 2) = 4x + 12 --> T(6x + 4) = 4x + 12 Now adding the two left-hand sides of the equations together, and the two right-hand sides. and using the additivity of T yields T(4) = -8x + 12 So 4 T(1) = -8x + 12. Thus, T(1) = -2x + 3 3. To get T(x^2) use the first equation you are given and the T(1) you just found. It'll turn out that T(x^2) = x^2 - x Now T(ax^2 + bx + c) = a T(x^2) + b T(x) + c T(1) which when you evaluate and collect like terms you get the answer that is given.
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