Verified answer

Let's work out the first few of f_n explicitly.

f_2(x) = f_1(f_1(x)) = f_1(x/2 + 10) = (x/2 + 10)/2 + 10 = x/4 + 15

In the same way (I'll leave you to confirm the details) you get

f_3(x) = x/8 + 17.5, f_4(x) = x/16 + 18.75, f_5(x) = x/32 + 19.375

It seems that the first part of each f_n is x/(2^n) and the second part is 20(1 - 1/(2^n))

This you can prove by induction. I'll leave that to you.

Then Limit f_n(x) = 20 because x/(2^n) will approach zero as n increases without limit (for any finite value of x) and the 1/(2^n) will also approach zero as n increases.

If there is a finite limit to your process, it will have to be a fixed point of f_1, coz f_1 is continuous.

Now f_1(x) = x iff x = 20.

Note that ( f_1(x) - 20) = (x - 20)/2. Hence f_n(x) - 20 = (f_(n-1)(x) - 20)/2.

So n ---> y_n = f_n(x) - 20 is a geometric sequence with 0 as its limit since | 1/2 | < 1.

I let you conclude.

good day! Does your instructor desire the respond written as a single fraction? if so, a million/a^2 + 6/a The least easy denominator is a^2. So, multiply the 2d fraction by potential of a/a. a million/a^2 + (6*a)/(a*a) a million/a^2 + 6a/a^2 Now that those 2 fractions have the comparable denominator, placed the numerators mutually. (a million + 6a)/a^2 very final answer: a million + 6a ------------ a^2 or 6a + a million ------------ a^2 desire this helps!