Need help, test question for tm

[MrRottenTreats]

hey guys, this is my first post.. ive been trying this question for litterally an hour and need some assistance.

' What point on the curve y=(lnx)³ is the tangent parrallel to the x-axis?'

ive tried many things from breaking the cubic up into:
y=(lnx)(lnx)(lnx)
y=ln(x+x+x)
y=ln(3x)
y'=1/3x (3)
y'=3/3x

but its not working out when you set y=0 to find the slope...

thank you for the help

 

[soroban]

Hello, MrRottenTreats!

Welcome aboard!

What point on the curve \(\displaystyle y\,=\,(\ln x)^3\) is the tangent parrallel to the x-axis?'

ive tried many things from breaking the cubic up into:
\(\displaystyle y\:=\\ln x)(\ln x)(\ln x)\)
\(\displaystyle y\:=\:ln(x\,+\,x\,+\,x)\) . . . . this is wrong

Why not differentiate immediately?

We have: \(\displaystyle \,y\:=\\ln x)^3\)

. . Then: \(\displaystyle \L\,y'\:=\:3(\ln x)^2\cdot\frac{1}{x}\)


If the tangent is parallel to the x-axis (horizontal), then \(\displaystyle y'\,=\,0\)

. . and we have: \(\displaystyle \L\,\frac{3(\ln x)^2}{x}\:=\:0\)

. . Therefore: \(\displaystyle \,\ln x\,=\,0\;\;\Rightarrow\;\;x\,=\,1\)

 

[MrRottenTreats]

i was jsut working this out, and you concluded lnx=0 therefore x=1 ..

then the point would be (1,O)

sorry i really want these right for my test tomarrow, need my mark to go up.

 

[dangerman77]

Yes:

And plugging this value back into the original:

So the point is, indeed, (1,0)

Good luck on that test!