13. Find a cubic function with the given zeros. 7, -3, 2
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We have to find the values of F.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.
Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1
As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40
Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
we see that :
x → 1 - 1/x → 1/(1-x) →x
so that:
g(g(g(x))) = x.
Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)
As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x
Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.
Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1
As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40
Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
we see that :
x → 1 - 1/x → 1/(1-x) →x
so that:
g(g(g(x))) = x.
Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)
As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x
Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.
<span>Ayesha's right. There's a good trick for knowing if a number is a multiple of nine called "casting out nines." We just add up the digits, then add up the digits of the sum, and so on. If the result is nine the original number is a multiple of nine. We can stop early if we recognize if a number along the way is or isn't a multiple of nine. The same trick works with multiples of three; we have one if we end with 3, 6 or 9.
So </span>
<span>has a sum of digits 31 whose sum of digits is 4, so this isn't a multiple of nine. It will give a remainder of 4 when divided by 9; let's check.
</span>
<span>
</span>Let's focus on remainders when we divide by nine. The digit summing works because 1 and 10 have the same remainder when divided by nine, namely 1. So we see multiplying by 10 doesn't change the remainder. So
has the same remainder as 
.
When Ayesha reverses the digits she doesn't change the sum of the digits, so she doesn't change the remainder. Since the two numbers have the same remainder, when we subtract them we'll get a number whose remainder is the difference, namely zero. That's why her method works.
<span>
It doesn't matter if the digits are larger or smaller or how many there are. We might want the first number bigger than the second so we get a positive difference, but even that doesn't matter; a negative difference will still be a multiple of nine. Let's pick a random number, reverse its digits, subtract, and check it's a multiple of nine:
</span>
So </span>
</span>
</span>Let's focus on remainders when we divide by nine. The digit summing works because 1 and 10 have the same remainder when divided by nine, namely 1. So we see multiplying by 10 doesn't change the remainder. So
When Ayesha reverses the digits she doesn't change the sum of the digits, so she doesn't change the remainder. Since the two numbers have the same remainder, when we subtract them we'll get a number whose remainder is the difference, namely zero. That's why her method works.
<span>
It doesn't matter if the digits are larger or smaller or how many there are. We might want the first number bigger than the second so we get a positive difference, but even that doesn't matter; a negative difference will still be a multiple of nine. Let's pick a random number, reverse its digits, subtract, and check it's a multiple of nine:
</span>