logorithmic and exponential functions

[astrostar]

Hi, got a question i'm a bit stuck with,

Plot the graph of y=e^-2x
Use the graph to find the gradient at x=0 (which i work out from the graph to be 1.8125). Does this sound correct and is there a way to work the gradient out by calculation to check my answer?

Also i'm asked to find the value of x when y=0.5 by calculation.
0.5=e^-2x
loge0.5=loge^-2x
loge0.5=x -2loge
loge0.5=x(x)-2
ln0.5=-2
ln0.5/-2= 0.35(2d.p) Does this seem right?

Any help appreciated. THnks

[Rassis]

Your first question:

If you mean the first derivative of y when x = 0, that is the slope of the line that passes the point whose coordinates are (x = 0; y = e^(-2 x 0) = 1) and is tangent to the function curve, then, you have dy/dx = -2.e^(-2.x). Making x = 0, you get dy/dx = -2.

Your second question:

0.5 = e^(-2.x)
ln(0.5) = -2.x.ln(e)
ln(0.5) = -2.x.1
x = -ln(0.5)/2
x = 0.34657

Rui

[astrostar]

Many thanks Rassis.

Astro

[Rassis]

My pleasure.

Rui