Melody

Hi Guest,

I do believe that Alan's answer is the correct one but I would like to understand what you have tried to do.

I think you have accidentally omitted an x in the second row but that is no bigf deal

When you say 'and with more bounding'  I have no idea what you mean.

That is a great solution Alan, which means I can easily understand it.  Thanks 

What real values of x are not in the domain of f(x) = 1/(abs(x^2 + 3x - 4) + abs(x^2 + 9x - 10))?

\(f(x) =\frac{ 1}{|x^2+3x-4|+ |x^2 + 9x - 10|}\\ f(x) =\frac{ 1}{|(x+4)(x-1)|+ |(x+10)(x-1)|}\\ x\ne 1 \)

Thanks Tiggsy,

Unfortunately, I do not understand how you have differentiated the RHS at all.   

Maybe you could explain a bit more please?

\((f'(x))^2 = f(x)f''(x)\\~\\ \frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}\\~\\ \int\frac{f'(x)}{f(x)}dx = \int\frac{f''(x)}{f'(x)}dx\\~\\ ln(f(x))+lnC=ln(f'(x))\qquad \text{Where lnC is a constant}\\~\\ ln(Cf(x))=ln(f'(x))\\~\\ Cf(x)=f'(x)\\~\\ C=\frac{f'(x)}{f(x)}\\~\\ \int C\;dx=\int \frac{f'(x)}{f(x)}\;dx\\~\\ Cx+k=ln(f(x))\\~\\ e^{Cx+k}=e^{ln(f(x))}\\~\\ f(x)=e^{Cx+k}\)

    Sub f(0)=1  and you get    k=0

\(f(x)=e^{Cx}\)

  \(​​​​f(x)=e^{Cx}\\ ​​​​f'(x)=Ce^{Cx}\\ ​​​​f''(x)=C^2e^{Cx}\\ ​​​​f'''(x)=C^3e^{Cx}\\ ​​​​f''''(x)=C^4e^{Cx}\\ \quad given\;\;f''''(0)=9\\ \quad \;C^4=9\\ \quad \;C=\pm\sqrt3\\ \)    

\(​​​​f'(x)=\sqrt3\;e^{\sqrt3\;x}\qquad or \qquad f'(x)=-\sqrt3\;e^{-\sqrt3\;x}\\ so\\ ​​​​f'(0)=\sqrt3\qquad or \qquad f'(0)=-\sqrt3\\\)

LaTex

(f'(x))^2 = f(x)f''(x)\\~\\
\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}\\~\\
\int\frac{f'(x)}{f(x)}dx = \int\frac{f''(x)}{f'(x)}dx\\~\\
ln(f(x))+lnC=ln(f'(x))\qquad \text{Where lnC is a constant}\\~\\
ln(Cf(x))=ln(f'(x))\\~\\
Cf(x)=f'(x)\\~\\
C=\frac{f'(x)}{f(x)}\\~\\
\int C\;dx=\int \frac{f'(x)}{f(x)}\;dx\\~\\
Cx+k=ln(f(x))\\~\\
e^{Cx+k}=e^{ln(f(x))}\\~\\
f(x)=e^{Cx+k}

​​​​f(x)=e^{Cx}\\
​​​​f'(x)=Ce^{Cx}\\
​​​​f''(x)=C^2e^{Cx}\\
​​​​f'''(x)=C^3e^{Cx}\\
​​​​f''''(x)=C^4e^{Cx}\\
\quad given\;\;f''''(0)=9\\
\quad \;C^4=9\\
\quad \;C=\pm\sqrt3\\

​​​​f'(x)=\sqrt3\;e^{\sqrt3\;x}\qquad or \qquad f'(x)=-\sqrt3\;e^{-\sqrt3\;x}\\
so\\
​​​​f'(0)=\sqrt3\qquad or \qquad f'(0)=-\sqrt3\\

If you would love to interact then try interacting with the other questions you have asked (that have been answered)

People want a response from you and they would like you to give them points for their time, effort and/or help.

I will take some time again with this question.. when I have time,  probably later today.

Tiggsy's answer is really good but I can give you a little more insight.

I just realized that this question is an AoPS question.  It is against their rules for you to post here,  

they want you to use their notice board and your own brain.

Sorry, We cannot help you.

The base of a solid is the area between the parabolas \(x=y^2 \) and \(2y^2 = 3 - x \) . Calculate hte volume of the solid given that the cross sections perpendicular to the -axis are equilateral triangles!

This is how I see it.

If  the y value at  a given point is k then the side length of the equilateral triangle will be 2k and the vertical height will be (sqrt3)k

The area of the triangle (the cross section) would be 0.5*2k*(sqrt3)k = (sqrt3)k^2

\(\displaystyle \int_0^1 \sqrt3*(\sqrt x)^2\;dx\;+\;\;\int_1 ^3 \sqrt3 *\left[\sqrt{\frac{3-x}{2}}\right]^2\;\;dx\\ =\displaystyle \int_0^1 \sqrt3*x\;dx\;+\;\;\int_1 ^3 \sqrt3 *\;\frac{3-x}{2}\;\;dx\\ =\frac{\sqrt 3}{2} \left[\displaystyle \int_0^1 2x\;dx\;+\;\;\int_1 ^3 \;(3-x)\;\;dx\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle \left[x^2 \right ]_0^1+\;\;\left[\;(3x-\frac{x^2}{2})\right]_1^3\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle1+\;\;\;(9-\frac{9}{2})-(3-\frac{1}{2})\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle1+\;\;\;(\frac{9}{2})-(\frac{5}{2})\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle3 \right]\\~\\ =\frac{3\sqrt3}{2}\;\;\;u^3 \)