Melody

\(f(x)=ax^4+bx^3+cx^2+dx+e\\ e=47\\ f(x)=ax^4+bx^3+cx^2+dx+47\\ f(1)=a+b+c+d+47=32\\ a+b+c+d=32-47\\ a+b+c+d=-15\\\)

Ths sum of the coefficients is -15

This would be true even if it the leading power was not 4.

What is the greatest integer x, where \(0\leq x\leq50\) that cannot be written as a(b+1)+b, where a and b are positive integers?

Well if the numbers are 1 and 24 then x will be 49.

I do not see how to get 50.  

So maybe the biggest x value is 49

sec x cot x sin^2 x

\(\sec x\cdot \cot x \cdot\sin^2 x\\ =\frac{1}{ \cos x}\cdot \frac{\cos x}{\sin x}\cdot \frac{\sin^2x}{1}\\ =\sin x\)

The rest cancels out.

That is a really helpful observation Alan. It makes the problem trivial. I'm impressed  :)

If  \(f(0)=f^{-1}(3a)\)  ,  solve for the value of a so that    \(f(0)=f^{-1}(3a)\)

This looks kinda interesting   -  I will assume that a is a constant.

\(let\;\\ y=\dfrac{a}{x+2} \quad x \ne-2\\ x+2=\dfrac{a}{y}\\ x=\dfrac{a}{y}-2\\ f^{-1}(x)=\dfrac{a}{x}-2\\~\\ f^{-1}(3a)=\dfrac{a}{3a}-2=\frac{1}{3}-2=\frac{-5}{3}\\ \)

\(f(0)=\frac{a}{2}\)

if

\(f(0)=f^{-1}(3a)\\ \frac{a}{2}=\frac{-5}{3}\\ a=\frac{-5*2}{3}\\ a=\frac{-10}{3}\\ a=-3\frac{1}{3} \)