{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.

Key Concepts

Fixed Points

Composition of Functions

Degree of a Polynomial

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