# 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?

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(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)# .

(a) There is some

(b)

(c)

(d) There is some

##### 1 Answer

#### Answer:

(a) and (d)

#### Explanation:

Let

Then

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

Then

So

So (a) is true.

Now let

Then

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

Then

So:

So (d) is true too.

**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

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.