Find the coordinates of the stationary point , S , on the curve.?

Question

Find the coordinates of the stationary point , S , on the curve y = (x+2)e^-x

Answer 1

y = (x + 2)e^(-x)

To find the coordinates of the stationary point (also known as local extrema), we must take the first derivative dy/dx and then make dy/dx = 0.

Using the product rule,

dy/dx = (1)e^(-x) + (x + 2)(-1)e^(-x)

Factor out e^(-x),

dy/dx = e^(-x) [ 1 + (x + 2)(-1) ]

dy/dx = e^(-x) [ 1 - (x + 2) ]

dy/dx = e^(-x) [ 1 - x - 2 ]

dy/dx = e^(-x) [-1 - x]

Now, set dy/dx = 0.

0 = e^(-x) [-1 - x]

Equate each factor to 0, and solve for the solutions.

e^(-x) = 0, -1 - x = 0

e^(-x) = 0 will have no real solutions; the other factor, however, will.

-1 - x = 0
-x = 1
x = -1

That makes our critical point (and potential stationary point) at x = -1.

Now, we test the behavior of dy/dx around x = -1. We want to test for positivity/negativity.
Make a number line consisting of our critical number, -1.

. . . . . . . . (-1) . . . . . . . .

Test x = -2: dy/dx = e^(-x) [-1 - x]
For x = -2, dy/dx = e^(-(-2)) [-1 - (-2)] = e^2 (1) = e^2
The result is positive, so on the number line, mark the region as positive.

. . .{+} . . . . (-1) . . . . . . . .

Test x = 0: dy/dx = e^(-0) [-1 - 0] = -1, which is negative. Mark the region as negative.

. . .{+} . . . . (-1) . . . .{-} . .

This tells us that the function is increasing from (-infinity, -1]
and decreasing from [-1, infinity).

It also tells us that we have a local maximum at x = -1.

When x = -1, y = (-1 + 2)e^(-(-1)) = 1(e)^1 = e, so
we have a local maximum at the point (-1, e).

(-1, e) is our stationary point S.

Answer 2

we take dy/dx which will
be 0 at the stationary point
d((x+2)e^-x)/dx
= -(x+2)*e^(-x)+1*e^(-x)
-e^(-x)(1+x)=0
e^(-x) is not a real solution
1+x=0
>>>x= -1 at stationary point
y=(-1+2)e^(-(-1))
=e

hence, coordinates of the
stationary point S on the
curve y = (x+2)e^-x
are (-1,e)

i hope that this helps

Answer 3

let u = (x+2) so du by dx = 1
let v= e^-x so dv by dx = -e^-x

(dv by dx * u) + (du by dx * v)
-e^-x(x+2) + e^-x = 0
-e^-x(x+2)=-e^-x
x+2= 0
x=2
sub in x value into orginal curve equation for y am sure ya can do that if ya askin questions like this
see ya

Answer 4

You find them.

Answer 5

Why do you even care? ...