Stationary Points - e^-3x

Show that curve y= e^-3x +2x has a stationary point at y=2/3(1+ln3/2)

how do i do this??

thanks

First you differentiate y= e^-3x +2x.

A stationary point has the property of having a gradient of 0.

So you make whatever you have as the differentiation of the function to equal 0. Then solve for x. Then sub in that value for x into y= e^-3x +2x.

Reply 2

McStewart

i got to

dy/dx = -3e^-3x + 2

e^3x = 3/2

but dont know where after that

Reply 3

Simba

maca

Show that curve y= e^-3x +2x has a stationary point at y=2/3(1+ln3/2)

how do i do this??

thanks

dy/dx = -3e^(-3x) + 2 = 0.

e^(-3x) = 2/3

-3x = ln(2/3)

x = -[ln(2/3)]/3

Putting this back into y = e^(-3x) + 2x:

y = e^[ln(2/3)] - 2[ln(2/3)]/3

= 2/3 + 2[ln(3/2)]/3

= 2[1 + ln(3/2)]/3.

Hope this helps,

~~Simba

Reply 4

McStewart

thanks, looks good to me!!!