EXPONENTIAL CURVE SKETCHING QUESTION (1 Viewer)

[jmadden]

QUESTION
For the curve,

find any stationary points, determine their nature and sketch the curve.

I found the derivative as

But I can't solve it for x when y' = 0 to find the stationary points.

HELP PLEASE.

 

[Aysce]

Derivative is 2xe^x^2 -2 not the one you found.

d/dx (e^x^2) = 2xe^x^2

 

[barbernator]

where did u get this question from? when finding stationary points, you are left with the equation x^2=-ln(x) and solving this is beyond the scope of mathematics

 

[Aysce]

where did u get this question from? when finding stationary points, you are left with the equation x^2=-ln(x) and solving this is beyond the scope of mathematics

What are you doing?

All you have to do is let the derivative = 0, find what x is, get the y-value and test for stat pt which gives you a minimum.

 

[Carrotsticks]

What are you doing?

All you have to do is let the derivative = 0, find what x is, get the y-value and test for stat pt which gives you a minimum.

This is what he was doing.

And solving it cannot be done by elementary methods.

 

[Aysce]

This is what he was doing.

And solving it cannot be done by elementary methods.

Aren't you able to do this:

2x(e^x^2)= 0, where e^x^2 =/= 0, and 2x = 0 where x=0 then just move on normally?

 

[RealiseNothing]

Aren't you able to do this:

2x(e^x^2)= 0, where e^x^2 =/= 0, and 2x = 0 where x=0 then just move on normally?

You still have the '-2' though don't you?

 

[Aysce]

The negative 2? During which step?

Because all im basically doing is solving 2x=0 and then just doing the normal stuff for an SP

 

[Parvee]

barbernator's method was right. You can't go any further.

 

[RealiseNothing]

The negative 2? During which step?

Because all im basically doing is solving 2x=0 and then just doing the normal stuff for an SP

You're finding when 2x(e^x^2) = 0

But the derivative is 2x(e^x^2) - 2 = 0

 

[Aysce]

Ohh I missed -2x... =.=

My fault, my fault.

 

[jmadden]

thanks for the help,
but what is the x value then found to be?

 

[barbernator]

thanks for the help,
but what is the x value then found to be?

approximately x=0.65, but you don't have to know this. It is beyond the scope of all HSC mathematics courses and the textbook shouldn't have that question