{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).

a. Without using a calculator, find the values of a, with 0 ≤ a ≤ 4, such that f has a fixed point. Give the fixed point in terms of a.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says consider the function G of X, which is equal to B multiplied by X, multiplied by the quantity of 2 minus X in quantity, where B is a real number and 1 is less than or equal to B, which is less than or equal to 3. Determine all fixed points of G in terms of B. We're given 4 possible choices as our answers. For choice A, we have X equal to 0. for choice B, we have X is equal to the quantity of 2 b minus 1 in quantity divided by B. For choice C, we have X equal to 0, and X is equal to the quantity of 2 b minus 1 in quantity divided by B. And for choice D, we have X equal to 0, and X is equal to the quantity of B minus 1 in quantity divided by B. Now as to determine all the fixed points for our function G. So, recall to find our fixed points, we need to set our function G of X. Equal to x. So that means that X is going to be equal to B multiplied by X, multiplied by the quantity of 2 minus X. So, we need to solve this equation for X. So the first step is going to be to subtract X from both sides of our equation. So we will have 0 is equal to. B multiplied by X, multiplied by the quantity of 2 minus X in quantity, minus X. Now, I can factor out an X out of both terms on the right-hand side of that expression. And so doing so, we'll have 0 is equal to X, multiplied by the quantity, and here we'll go ahead and distribute our B, so we'll have 2 B minus B multiplied by X minus 1. And so now the software X will just set each of these factors equal to 0 and solve for X. So for the first factor, we'll just have X equal to 0, and for the second factor, we'll have 2 B minus B multiplied by X, minus 1 is equal to 0. So solving this expression for X, the first thing we'll need to do is to add the quantity of 1 minus 2B to both sides to this of this equation, and we will get minus B, multiplied by X is equal to 1 minus 2B. And now we'll just divide both sides of the equation by minus B and we will get X is equal to the quantity of 2 B minus 1 in quantity divided by B. So we have found two fixed points, one at X equal to 0 and one at X equal to the quantity of 2b minus 1 in quantity divided by B. And if I look at my answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

Watch next