Let f be a continuous function defined on the interval [0,1] such that f(0) = f(1). show that there exists a number 0 < = a < = 1/2 such that f(a) = f(a+1/2)

Explanation:

In order to prove that affirmation, we define the function g over the interval [0, 1/2] with the formula

If we evaluate g at the  endpoints we have

g(0) = f(1/2)-f(0) = f(1/2) - f(1) (because f(0) = f(1))  

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

Since g(1/2) = -g(0), we have one chance out of three

We will prove that g has a zero on [0,1/2]. If g(0) = 0, then it is trivial. If g(0) ≠ 0, then we are in one of the first two cases, and therefore g(0) * g(1/2) < 0. Since f is continuous, so is g. Bolzano's Theorem assures that there exists c in (0,1/2) such that g(c) = 0.  This proves that g has at least one zero on [0,1/2].

Let c be a 0 of g, then we have

Hence, f(c+1/2) = f(c) as we wanted.

i use my brain for prombles like this maybe you could try it

step-by-step explanation:

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.