{Use of Tech} Fixed points of quadratics and quartics Let f(x)...

Question

{Use of Tech} Fixed points of quadratics and quartics Let f(x) = ax(1 -x), where a is a real number and 0 ≤ a ≤ 1. Recall that the fixed point of a function is a value of x such that f(x) = x (Exercises 48–51).

b. Consider the polynomial g(x) = f(f(x)). Write g in terms of a and powers of x. What is its degree?

b. Consider the polynomial g(x) = f(f(x)). Write g in terms of a and powers of x. What is its degree?

Accepted Answer

Welcome back, everyone. Consider the function G of X equals BX multiplied by 2 minus X, where B is a real number and B is between 1 and 3 inclusive. If we define a new function p of X equals G of G of X, what is the degree of PF X? We're given 4 answer choices A says 3, B4, C2, and D1. So we're looking at the composition G of G of X, which essentially means that for every X that we have in G of X, we are now replacing it with G of X, right? So we're going to get B multiplied by G of X instead of X, which is then multiplied by 2 minus G of X, right, because now we are replacing the independent variable X with G of X. And now if we replace G of X with its definition, we're going to get B multiplied by BX2 minus X, multiplied by 2 minus BX multiplied by 2 minus X. So this is our new function p of X, and what we want to do is simply understand what's the degree of this function. So, We are simply going to distribute. First of all, we can say that PF X is equal to B2 X. Multiplied by 2 minus X, multiplied by 2 minus BX, multiplied by 2 minus X, and from here we can start distributing. So first of all, We can multiply by 2 and we're going to get 2 B 2 X. Multiplied by 2 minus X, and then we are multiplying by the second term, right, which gives us minus B cubed X squared multiplied by 2 minus X2. And now let's go ahead and distribute further. So for the first part, we're going to get 4b2 X minus 2B2X2. Now we have to square 2 minus X, and here we can apply the formula. So we have a difference of two terms. We are squaring that difference, so we're squaring the first term that gives us 4. We're subtracting the double product of the two terms, so 2 X multiplied by 2 is 4 X, and then we're adding the square of the second term, so we're adding X squared. So we're basically taking B cubed X squared multiplied by. 4 minus 4 x plus x squad, and then we can distribute, right? So we're going to get. Negative or be cubed X2 plus because we're multiplying by a negative number. So in this case we're adding 4 B cubed X cubed, and then we are subtracting B cubed X to the power of 4th. And now what we want to do is simply identify. The X with the highest exponents. So we have X to the 1st, X2 X2 gain x cubed, and X to the power of 4, and this is the highest. Exponent on the independent variable x. Not that we're ignoring the B terms because they are constants, right? So essentially our answer to this problem corresponds to the answer choice B4. Thank you for watching.

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