Question

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 [0, 1]` such that `f(x+1) = f(x)`.

(b) `f(2-x) = f(x)` for all `x in [0, 2]`.

(c) `f(x) <= 1` for all `x in [0, 2]`.

(d) There is some `x in [0, 2]` such that `f(x) = -f(2-x)`.

Answer 1

(a) and (d)

Explanation:

Let `g(x) = f(x+1)-f(x)`

Then `g(x)` is defined and continuous on the interval `[0, 1]`

We find:

`g(0) = f(1)-f(0) = 1-(-1) = 2`

`g(1) = f(2)-f(1) = -1-1 = -2`

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

Then `f(y+1) - f(y) = g(y) = 0`

So `f(y) = f(y+1)`

So (a) is true.

Now let `h(x) = f(x) + f(2-x)`

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

We find:

`h(0) = f(0) + f(2) = -1+(-1) = -2`

`h(1) = f(1) + f(1) = 1+1 = 2`

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

Then `f(y)+f(2-y) = h(y) = 0`

So: `f(y) = -f(2-y)`

So (d) is true too.

`color(white)()`
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)`

`color(white)(f(x)) = 1/3(-8x^4 + 40x^3 - 70x^2 + 44x - 3)`

Then:

`f(0) = -1`

`f(1/2) = 1/3(-1/2+5-35/2+22-3) = 2`

`f(1) = 1/3(-8+40-70+44-3) = 1`

`f(3/2) = 1/3(-81/2+135-315/2+66-3) = 0`

`f(2) = 1/3(-128+320-280+88-3) = -1`

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

Note that:

`f(1/2) = 2 != 0 = f(3/2) = f(2-1/2)" "` so (b) fails.

Note that:

`f(1/2) = 2 > 1" "` so (c) fails.