Which function is this?

Algebra Level 4

Let $f: \mathbb{R} \rightarrow [\frac{1}{2}, 1 ]$ and $f\left(x+2\right)= \frac{1}{2} +\sqrt{f\left(x\right)-f\left(x\right)^2}$ Then which among the following is always true?

At least 2 of the given choices

f(2) = f(4)

f(3) = f(7)

f(4) = f(10)

3 solutions

Plinio Sd
Jul 17, 2015

A possible set of values of $f$ that satisfies this equation is, for $n \in \mathbb{N}$ , $f(4n) = 1 \\ f(4n+2) = \dfrac{1}{2}$ This already excludes all options, except the true one.

But, let us show why the remaining option is always true. If we take the square at both sides of the equation we find, for $x \in \mathbb{R}$ , $f(x+2)^2 = \dfrac{1}{4} + \sqrt{f(x)-f(x)^2} + f(x) - f(x)^2,$ which is always true, since the terms are positive. Using the equation again to substitute the square root, it give us $f(x+2)^2 = \dfrac{1}{4} + f(x+2) - \dfrac{1}{2} + f(x) - f(x)^2,$ then, rearranging the above equation, we find $f(x) - f(x)^2 = \dfrac{1}{4} - (f(x+2) - f(x+2)^2).$ Using the above equation in itself, by substituting the right-hand side term, we find $f(x) - f(x)^2 = f(x+4) - f(x+4)^2.$ Finally, we can solve that for $f(x)$ and determine the unique result $f(x) = f(x+4)$ , since the image of the function is in the interval $\left[\dfrac{1}{2}, 1\right]$ .

Well, for your example, you do have to specify what $f$ is on all real numbers, not just even integers. But this is a minor nitpick.

Patrick Corn - 5 years, 11 months ago

Yeah, you are right. I was not very clear about that. Thanks for the remark.

Plinio SD - 5 years, 11 months ago

(f(x+2)-1/2)^2 = f(x)-f(x)^2 f(x+2)^2-f(x+2)+1/4 = f(x)-f(x)^2 f(x=x+2)-f(x=x+2)^2 = f(x+4)^2-f(x+4)+1/4 1/4+f(x+2)^2-f(x+2) = f(x)-f(x)^2 = f(x+4)^2-f(x+4) from this we can see that f(x+2) = 1/2 + sqrt(f(x)-f(x)^2)=1/2 + sqrt(f(x+4)^2-f(x+4)) so we need f(x) and f(x+4) to equate so f(3) = f(7)

Joey Chemis - 5 years, 10 months ago

Hadia Qadir
Aug 4, 2015

The correct answer is f(3) = f(7). There is a deal of tedious elementary algebraic manipulation of the given equation to arrive at the conclusion that f(x) = f(x+4), but the job is doable by a high school student. Problems of this sort were assigned to my honors Precalculus classes. Have fun.

Vineet Singh
Jul 18, 2015

[f(x+2)- 1/2]^2 = 1/4 - [f(x) - 1/2]^2 putting x=x+2
[f(x+4)- 1/2]^2 = 1/4 - 1/4 +[f(x) - 1/2]^2 = [f(x) - 1/2]^2 => f(x+4) = f(x).

Which function is this?

3 solutions

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