# Suppose f(x) is a real-valued function, defined and continuous on [0, 2] with f(0) = -1, f(1) = 1 and f(2) = -1. Which of the following statements are true?

## (a) There is some $x \in \left[0 , 1\right]$ such that $f \left(x + 1\right) = f \left(x\right)$. (b) $f \left(2 - x\right) = f \left(x\right)$ for all $x \in \left[0 , 2\right]$. (c) $f \left(x\right) \le 1$ for all $x \in \left[0 , 2\right]$. (d) There is some $x \in \left[0 , 2\right]$ such that $f \left(x\right) = - f \left(2 - x\right)$.

Apr 15, 2017

(a) and (d)

#### Explanation:

Let $g \left(x\right) = f \left(x + 1\right) - f \left(x\right)$

Then $g \left(x\right)$ is defined and continuous on the interval $\left[0 , 1\right]$

We find:

$g \left(0\right) = f \left(1\right) - f \left(0\right) = 1 - \left(- 1\right) = 2$

$g \left(1\right) = f \left(2\right) - f \left(1\right) = - 1 - 1 = - 2$

So by Bolzano's theorem there is some $y \in \left(0 , 1\right)$ such that $g \left(y\right) = 0$

Then $f \left(y + 1\right) - f \left(y\right) = g \left(y\right) = 0$

So $f \left(y\right) = f \left(y + 1\right)$

So (a) is true.

Now let $h \left(x\right) = f \left(x\right) + f \left(2 - x\right)$

Then $h \left(x\right)$ is defined and continuous on the interval $\left[0 , 2\right]$, in particular on the interval $\left[0 , 1\right]$

We find:

$h \left(0\right) = f \left(0\right) + f \left(2\right) = - 1 + \left(- 1\right) = - 2$

$h \left(1\right) = f \left(1\right) + f \left(1\right) = 1 + 1 = 2$

So by Bolzano's theorem there is some $y \in \left(0 , 1\right)$ such that $h \left(y\right) = 0$

Then $f \left(y\right) + f \left(2 - y\right) = h \left(y\right) = 0$

So: $f \left(y\right) = - f \left(2 - y\right)$

So (d) is true too.

$\textcolor{w h i t e}{}$
Bonus

Let:

f(x) = -1/(0!)+3/(1!)(2x)-4/(2!)(2x)(2x-1)+4/(3!)(2x)(2x-1)(2x-2)-4/(4!)(2x)(2x-1)(2x-2)(2x-3)

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{3} \left(- 8 {x}^{4} + 40 {x}^{3} - 70 {x}^{2} + 44 x - 3\right)$

Then:

$f \left(0\right) = - 1$

$f \left(\frac{1}{2}\right) = \frac{1}{3} \left(- \frac{1}{2} + 5 - \frac{35}{2} + 22 - 3\right) = 2$

$f \left(1\right) = \frac{1}{3} \left(- 8 + 40 - 70 + 44 - 3\right) = 1$

$f \left(\frac{3}{2}\right) = \frac{1}{3} \left(- \frac{81}{2} + 135 - \frac{315}{2} + 66 - 3\right) = 0$

$f \left(2\right) = \frac{1}{3} \left(- 128 + 320 - 280 + 88 - 3\right) = - 1$

Note that since $f \left(x\right)$ is a polynomial it is continuous everywhere, including on the interval $\left[0 , 2\right]$.

Note that:

$f \left(\frac{1}{2}\right) = 2 \ne 0 = f \left(\frac{3}{2}\right) = f \left(2 - \frac{1}{2}\right) \text{ }$ so (b) fails.

Note that:

$f \left(\frac{1}{2}\right) = 2 > 1 \text{ }$ so (c) fails.