The miniumim distance from the origin to the curve y = e^x is?

Question

please show work for best answer!!!

Answer 1

d=sqrt( x^2 +e^2x)
d'=0 you'll get x+e^2x=0
solve for x
x=-0.426
d=0.78

Answer 2

The origin is located at (0, 0).
We want to find the minimum distance between (0, 0) and the curve y = e^x.

The distance formula goes as follows:

D = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 )

The distance between (0, 0) and the curve y = e^x is calculated in the following manner:

D = sqrt ( (x - 0)^2 + (y - 0)^2 )

But y = e^x, so

D = sqrt ( (x - 0)^2 + (e^x - 0)^2 )

Which simplifies to

D = sqrt ( x^2 - (e^x)^2 )

To minimize this distance, we calculate dD/dx and then make it 0. But to simplify things, we can square both sides and differentiate implicitly.

D^2 = x^2 - (e^x)^2
D^2 = x^2 - e^(2x)

Differentiating,

2D (dD/dt) = 2x - e^(2x) (2)

Making dD/dt = 0,

2D (0) = 2x - 2e^(2x)

0 = 2x - 2e^(2x)

0 = x - e^(2x)

Solve for x (which isn't an easy task), those will be critical numbers.

Answer 3

d^2 = x^2 +e^2x
d^2´= 2x+2e^2x = 0 you have to solve
z+2 e^z=0 with z= 2x you´ll find z= -0.85261 and d^2=0.6080368