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OlgaM077 [116]
3 years ago
7

Let T:P3→P3 be the linear transformation such that T(â’2x2)=3x2+4x, T(0.5x+3)=â’3x2+3xâ’4, and T(2x2â’1)=â’3x+4. Find T(1), T(

x), T(x2), and T(ax2+bx+c), where a, b, and c are arbitrary real numbers.
Mathematics
1 answer:
balu736 [363]3 years ago
8 0

It looks like we're told that

T(-2x^2)=3x^2+4x

T\left(\dfrac12x+3\right)=-3x^2+3x-4

T(2x^2-1)=-3x+4

We use the fact that T is linear to find T(1),T(x),T(x^2). First, we notice that

T(-2x^2)+T(2x^2-1)=T(-2x^2+2x^2-1)=T(-1)=-T(1)

We also have

T\left(\dfrac12x+3\right)=T\left(\dfrac12x\right)+T(3)=\dfrac12T(x)+3T(1)

T(2x^2-1)=T(2x^2)-T(1)=2T(x^2)-T(1)

So once we find T(1), we can determine T(x) and T(x^2). We have

T(1)=-\left(T(-2x^2)+T(2x^2-1)\right)\implies T(1)=-3x^2-x-4

and using this we find

T(x)=12x^2+12x+16

T(x^2)=-\dfrac32x^2-2x

Then

T(ax^2+bx+c)=T(ax^2)+T(bx)+T(c)=aT(x^2)+bT(x)+cT(1)

T(ax^2+bx+c)=-\dfrac32\left(a-8b+2c\right)x^2-(2a-12b+c)x+16b-4c

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