Solving Linear Algebra: f(5x5+4x2+3) & 1-1 & Onto

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So in my linear class we are reveiwing some things befor we start working out of the book and i got this as a practice ?. I am confused on what is being asked and how to start off.


Let P5 be the set of all polynomials of degree 5 or less. Let p(x) be an element of P5 with p'(x) its derivitive. define function f as follows.

f: P5\rightarrowP5

f(p(x))=xp'(x) + 1

a. find f(5x5+4x2+3)

b. is f 1 to 1 and explain

c. is f onto and explain
 
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In Linear Algebra, you often have to start off with definitions themselves. Once you have the definitions in hand, the proofs become considerably easier.

What does 1-1 mean? What does onto mean? These are definitions. 1-1 means that f(x) = f(y) implies that x = y. This essentially means that no two unequal points in the domain map to the same point in the range. Onto means that every point in the range is taken up. For example, if we define f: R to R by f(x) = 2, then f is not onto because there is no point x0 in the domain such that f(x0) = 3, for instance. After you learn these definitions, the questions should be easy to answer.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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