Turning Points

[Anthonyk2013]

I know this is basic algebra but cant remember how to find X, Is my work correct so far?

Finding turning points

Y=5X-2lnX

dy/dx=5-2/x

dy/dx=5-2x^-1
dydx=0

5-2x^-1 =0
-2x^-1=-5

 

[Anthonyk2013]

I know this is basic algebra but cant remember how to find X, Is my work correct so far?

Finding turning points

Y=5X-2lnX

dy/dx=5-2/x

dy/dx=5-2x^-1
dydx=0

5-2x^-1=0
-2x^-1=-5

Think I sorted this

5-2/x=0

-2/x=-5

5x=2

x=2/5

x=0.4

 

[Ishuda]

Think I sorted this

5-2/x=0

-2/x=-5

5x=2

x=2/5

x=0.4

Correct
EDIT: I looked at this as just a solution to the 5-2/x=0 part and was not looking at the turning point part. As Subhotosh Khan pointed out, d²y/dx² also needs to be non-zero [or, if it is, then the first non-vanishing derivative need to be an even derivative]

 

[Deleted member 4993]

Think I sorted this

5-2/x=0

-2/x=-5

5x=2

x=2/5

x=0.4

dy/dx = 0 does not always give you turning points (i.e. it is a necessary condition but NOT sufficient).

You need to also check \(\displaystyle \frac{d^2y}{dx^2}\) and show that \(\displaystyle \frac{d^2y}{dx^2} \ \ne \ \ 0\) at the point you are claiming to be turning point.