1. Graph this function

Graph: $\displaystyle f(x)=3(1-e^{-x})$, for nonnegative values of x.

Of course I could plug in numbers, but I think they want me to employ some properties of logarithms to make graphing easier.

I see the vertical stretching of 3 and horizontal translation of -1. I believe the -x will flip it across the y-axis... Any Ideas?

2. Originally Posted by ChristopherDunn
Graph: $\displaystyle f(x)=3(1-e^{-x})$, for nonnegative values of x.

Of course I could plug in numbers, but I think they want me to employ some properties of logarithms to make graphing easier.

I see the vertical stretching of 3 and horizontal translation of -1. I believe the -x will flip it across the y-axis... Any Ideas?

Let $\displaystyle y=f(x)$. Logarithms won't help much because of the $\displaystyle \ln(1-e^{-x})$ term which wouldn't help much.

IMO the best way to go about it is to find the points of intersection with the axes and any asymptotes after distributing the 3 and then draw a curve similar to $\displaystyle -\frac{1}{x}$

Spoiler:
For example we know that as $\displaystyle x \rightarrow \infty$ then $\displaystyle e^{-x} \rightarrow 0$ so there will be an asymptote at $\displaystyle y=3$

Similarly to find the point of intersection on the x axis solve f(0)

$\displaystyle f(0) = 3(1-1) = 0$ hence (0,0) is a point.

You can then transform the graph left by 1 unit.

Note how the graph below is similar to $\displaystyle y=1-\frac{1}{x}$ which is shown in green

edit: on the graphs I have restricted the domain to $\displaystyle x>0$ as that is what your question states

edit 2: I'm a fool, I misunderstood what you meant by the -x (I think)

3. hhmmmm I think I got it a different way...

let $\displaystyle f(x)=e^x$
$\displaystyle f(-x)=e^{-x}$
$\displaystyle -f(-x)=-e^{-x}$
$\displaystyle -f(-x)+1=-e^{-x}+1$
$\displaystyle 3(-f(-x)+1)=3(-e^{-x}+1)$

reducing to $\displaystyle -3(f(-x)-1)$

so I can graph $\displaystyle e^x$, and then working from the inside out, flip it across the y axis (-x), translate vertically down one (-1), then flip over x-axis and stretch by 3 (-3). That all seems more complicated than picking points, though....

I had had originally thought that they wanted me to use logs because that is what the chapter is about... but it is also about exponential functions.