Melody

10000000000004030

10*  1 000 000 000 000 403

= 2 * 5 *  1 000 000 000 000 403

That is it.

f(x)=x^(101) + Ax + B

x^2-1=(x-1)(x+1)  so  1 and -1 are roots of f(x)

so

f(1)=0    and  f(-1)=0

solve simultaneously.

Easy peazy.

The largest value would have to be 2.  I got that intuitively.

Then I set out to prove it.

Let 

 \(z=a+bi\\ |z|=\sqrt{a^2+b^2}=1\\ a^2+b^2=1\\\)

\(z^2+1 = a^2-b^2+2abi+1\\ z^2+1 = a^2-b^2+1+2abi\\ |z^2+1| =\sqrt{ (a^2-b^2+1)^2+(2ab)^2}\\ |z^2+1| =\sqrt{ (a^2+b^2-2b^2+1)^2+(2ab)^2}\\ |z^2+1| =\sqrt{ (1-2b^2+1)^2+(2ab)^2}\\ |z^2+1| =\sqrt{ (2-2b^2)^2+(2ab)^2}\\ |z^2+1| =\sqrt{ (4+4b^4-8b^2)+(2ab)^2}\\ |z^2+1| =2\sqrt{ (1+b^4-2b^2)+(ab)^2}\\ |z^2+1| =2\sqrt{ (1-b^2)^2+(ab)^2}\\ |z^2+1| =2\sqrt{ (a^2)^2+(ab)^2}\\ |z^2+1| =2a\sqrt{ (a^2)+(b)^2}\\ |z^2+1| =2a\sqrt{ 1}\\ |z^2+1| =2a\\ \text{But a must be between 0 and 1}\\ \text{So the largest possible value of } |z^2+1| \;\;is\;\;2\)

there is probably a much better way to do it. 

Danke Probolobo,    

You are very welcome.

infinity is a concept, not a number.  You cannot do maths on it like that.

\(\displaystyle \lim_{x\rightarrow \infty}1^{x*-1}\\ =\displaystyle \lim_{x\rightarrow \infty}1^{-x}\\ =1\)

You are welcome,  even a dad's help need to be checked  LOL

Your notation is fine Jenny,     (although brackets do cut down any possibility of confusion) 

f(x)=3/5x^3-9

\(f(x)=\frac{3}{5}x^3-9\\ let \;\;y=f(x)\\ y=\frac{3}{5}x^3-9\\ \text{make x the subject}\\ 5y=3x^3-45\\ \frac{5y+45}{3}=x^3\\ x=\sqrt[3]{\frac{5y+45}{3}}\\ so\\ f^{-1}(x)=\sqrt[3]{\frac{5x+45}{3}}\\\)

I don't know what they want for the verification ....

Here is graph validation I guess

A function and its inverse are reflections of one another over the line y=x

Latex

f(x)=\frac{3}{5}x^3-9\\
let \;\;y=f(x)\\
y=\frac{3}{5}x^3-9\\
\text{make x the subject}\\
5y=3x^3-45\\
\frac{5y+45}{3}=x^3\\
x=\sqrt[3]{\frac{5y+45}{3}}\\
so\\
f^{-1}(x)=\sqrt[3]{\frac{5x+45}{3}}\\

I have answered and I do expect to get a response from you. (on the question thread)

Dave had some 50-cent coins. Wayne had some $1 coins. Dave gave 50% of his coins to Wayne and Wayne gave 50% of his coins to Dave. Wayne used some of the 50-cent coins to buy a book that cost $5. Dave used some of his $1 coins to buy a bag that cost $16. After that the ratio of the number of 50-cent coins to $1 coin Dave had was 1:2 and the number of 50-cent coins to the number of $1 coins Wayne had was 1:5. How much did Dave have at first?

slight simplification:   

Dave had some 50-cent coins.     I let that be  2x  (50c) coins   

Wayne had some $1 coins.          I let that be  2y  ($1) coins   

Dave gave 50% of his coins to Wayne and Wayne gave 50% of his coins to Dave.

Now they each have    x (50c)  +  y ($1)

Wayne used some of the 50-cent coins to buy a book that cost $5.  So wayne spends  10 of his 50c peices

Dave used some of his $1 coins to buy a bag that cost $16.             So Dave spends  16 of his $1 coins

So now 

Wayne has   (x - 10)  (50c coins)   +  y  ($1coins)                 which means that Wayne has   $  \((\frac{x-10}{2}+y)\)

Dave has          x       (50c coins)   +  (y-16)    ($1coins)        which means that Dave has      \($(\frac{x}{2}\;+\;\;y-16)\)

After that the ratio of the number of 50-cent coins to $1 coin Dave had was 1:2

\(\frac{x}{y-16}=\frac{1}{2}\\ 2x=y-16\\ y=2x+16\\\)

and the number of 50-cent coins to the number of $1 coins Wayne had was 1:5. 

\(\frac{x-10}{y}=\frac{1}{5}\\ y=5x-50\)

Solve simultaneously

\(2x+16=5x-50\\ 66=3x\\ x=22\\ y=44+16=60\)

How much did Dave have at first?  

He had 2x   50c coins so initially he had    44  fifty cent peices    =    $22

Check:

Started with Dave having 44  fifties and Wayne having 120 dollars

Dave

44 fifties

22 fifties and 60 ones

22 fifties and 44 ones         Ratio 1 to 2 good

Wayne

120 dollars

22 fifties and 60 ones

12 fifties and 60 ones        Ratio  1 to 5  good