3
$\begingroup$

I am working on the question below and I am getting stuck.

Consider the surge function $y=axe^{-bx}$ with $a$ and $b$ positive constants.

(a) Find the local maxima, local minima, and inflection points.

(b) How does varying $a$ and $b$ affect the shape of the graph?

(c) On one set of axes, sketch the graph of this function for a few values of $a$ and $b$.

I've played around with the function on Wolfram Alpha in order to get a feel for how different values of $a$ and $b$ affect the graph. I also see that the local maximum seems always to be at $1/b$. I cant seem to figure out how to show this result using first and second derivatives, etc.

I've found the first derivative to be: $$y'=ae^{-bx}(1-b)$$

  • 4
    $y'=ae^{-bx}-baxe^{-bx} = ae^{-bx}(1-bx)$.2011-11-17
  • 0
    Oops, yes I see that now, I had not factored it properly. So $(1-bx)=0$, which leads to $x=1/b$.2011-11-17
  • 0
    Assuming $a>0$ and $b>0$ then all the possible curves are similar. You have $\left(\frac b a\right) y = (bx)e^{-bx}$, so increasing $a$ while leaving $b$ unchanged stretches the curve away from the $x$-axis (i.e. increases the magnitude of $y$ while leaving $x$ unchanged), while increasing $a$ and $b$ by the same proportionate amount shrinks the curve towards the $y$-axis (i.e. reduces the magnitude of $x$ while leaving $x$ unchanged). So increasing $b$ without changing $a$ would shrink towards the origin. Negative $a$ or $b$ would just flip the curve across the axes2018-07-25

1 Answers 1

0

From David Mitra:

$$y^\prime =ae^{−bx}−baxe^{−bx}=ae^{−bx}(1−bx)$$

From NehriMattisse:

So $(1−bx)=0$, which leads to $x=1/b$.