Lesson Evaluating a function defined by functional equation
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<H2>Evaluating a function defined by functional equation</H2> <H3>Problem 1</H3>The function f satisfies {{{f(sqrt(2x - 1))}}} = {{{1/(2x - 1)}}} for all x not equal to 1/2. Find f(2). <B>Solution</B> The idea is to find x such that {{{sqrt(2x-1)}}} = 2 and then calculate f(2) using the given functional equation. Below is an implementation of this idea in steps. <pre> <U>Step 1.</U> We want to find x from equation {{{sqrt(2x-1)}}} = 2. Do all necessary transformations as follows 2x-1 = 2^2 = 4, 2x = 4 + 1 = 5, x = 5/2. <U>Step 2</U>. According to the functional equation, {{{f(sqrt(2x - 1))}}} = {{{1/(2x - 1)}}}. Substitute here x = 5/2. Remember that x is determined in a way that {{{sqrt(2x-1)}}} = 2. Therefore, you will get f(2) = {{{1/(2*(5/2)-1)}}} = {{{1/(5-1)}}} = {{{1/4}}}. At this point, the problem is solved completely. <U>ANSWER<</U>. f(2) = {{{1/4}}}. </pre> <H3>Problem 2</H3>Suppose a function f is such that f(1/x) - 3f(x) = x for every non-zero x. Find f(2). <B>Solution</B> <pre> Write the given functional equation for x = 2 and for x = 1/2. For x = 2, you will have f(1/2) - 3f(2) = 2 (1) For x = 1/2, you will have f(1/(1/2)) - 3f(1/2) = 1/2, or f(2) - 3f(1/2) = 1/2. (2) Let u = f(1/2), v = f(2) for breavity. Then you can re-write (1) and (2) in the form u - 3v = 2 (3) v - 3u = 1/2 (4) Now I want to solve the system of linear equations (3) and (4) and find v. For it, I express u = 2+ 3v from (3) and substitute it into equation (4) v - 3*(2+3v) = 1/2. In the last equation, multiply both sides by 2 to run from denominator 2v - 6*(2+3v) = 1 Simplify and find v 2v - 12 - 18v = 1 -16v = 1 + 12 -16v = 13 v = {{{-13/16}}}. Thus v = f(2) = {{{-13/16}}}, and the problem is just solved. <U>ANSWER</U>. f(2) = {{{-13/16}}}. </pre> <H3>Problem 3</H3>If {{{f(3x/(x-4))}}} = {{{x^2 + x + 1}}}, what is the value of f(5)? <B>Solution</B> <pre> To solve the problem, find the value of x such that {{{(3x)/(x-4)}}} = 5. It is easy: first, cross multiply to get 3x = 5(x-4), then simplify step by step 3x = 5x - 20 20 = 5x - 3x 20 = 2x ---> x = 20/2 = 10. Now substitute x= 10 into the given formula. Since {{{(3x)/(x-4)}}} = 5 at x= 10, you will get then f(5) = {{{10^2 + 10 + 1}}} = 100 + 10 + 1 = 111. <U>ANSWER</U>. Under imposed conditions, f(5) = 111. </pre> <H3>Problem 4</H3>If f(x) = 1/(1 - x) , find (f(f(f(f...f)(sqrt2), (45 times). <B>Solution</B> <pre> If f(x) = {{{1/(1-x)}}}, then f(f(x)) = {{{1/(1-1/(1-x))}}} = {{{(1-x)/((1-x)-1)}}} = {{{(x-1)/x}}}, then f(f(f(x))) = {{{1/(1-(x-1)/x)}}} = {{{x/(x-x+1)}}} = x. Thus, applying function f to any real number x =/= 1, x =/= 0 three times, we get the value of x again. In other words, f(f(f(x))) == x identically, for all real x =/= 1, x =/= 0. So, for example, {{{f(f(f(sqrt(2))))}}} = {{{sqrt(2)}}}; {{{f(f(f(sqrt(3))))}}} = {{{sqrt(3)}}}; {{{f(f(f(sqrt(5))))}}} = {{{sqrt(5)}}}; {{{f(f(f(sqrt(7))))}}} = {{{sqrt(7)}}}; {{{f(f(f(root(3,2))))}}} = {{{root(3,2)}}}; {{{f(f(f(root(3,3))))}}} = {{{root(3,3)}}}; {{{f(f(f(root(3,5))))}}} = {{{root(3,5)}}}; {{{f(f(f(root(3,7))))}}} = {{{root(3,7)}}}, and so on. Since 45 is a multiple of 3, f applied to {{{sqrt(2)}}} 45 times is {{{sqrt(2)}}}; f applied to {{{sqrt(3)}}} 45 times is {{{sqrt(3)}}}; f applied to {{{sqrt(5)}}} 45 times is {{{sqrt(5)}}}; f applied to {{{sqrt(7)}}} 45 times is {{{sqrt(7)}}}; f applied to {{{root(3,2)}}} 45 times is {{{root(3,2))}}}; f applied to {{{root(3,3)}}} 45 times is {{{root(3,3))}}}; f applied to {{{root(3,5)}}} 45 times is {{{root(3,5))}}}; f applied to {{{root(3,7)}}} 45 times is {{{root(3,7))}}}, and so on. </pre> Solved and significantly expanded. For example, f applied 2025 times to the number {{{root(2025,2025))}}} is {{{root(2025,2025)}}}. Similarly, f applied 2025 times to the number {{{2025^2025}}} is {{{2025^2025}}}. As well as f applied 2025 times to the number 2025! is 2025! , again. You can easily construct a million other examples. 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