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)#.

1 Answer
Apr 15, 2017

Answer:

(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.