How to plot the exponential function?

Question

Why do we, in the exponential function $a^x$, take $a>0$ and $a\not=0$ and how can we plot the graph of the exponential function $a^x$?

Answer 1

If $a<0$, we run into complex numbers and that complicates things a bit. If $a=0$, the function becomes rather boring, since $0^x=0$ for all $x>0$ and undefined (over the Reals) for $x\leq 0$.

As for how to plot it, you can simply type it into your calculator, or Wolfram Alpha, or whatever your preferred tool is, or you could calculate a few points by hand and connect the dots.

Answer 2

At first $a\neq 0$ because if a=0 then the range of the function is a Singleton set {0}.And if a<0 then the function is not continuous at all.

Let us examine this statement more precisely.if a is negative then $a^x$ have a positive value when x is even and negative value when x is odd. Therefore a must not be negative if you say that the function is continuous.

For graphing the function you can use the intermediate value theorem. Note the the domain of the function is $(-\infty,\infty) $.when x→-$\infty$ ,$a^x$→0 . and when x→$\infty$ ,$a^x$→$\infty$ therefore the graph looks like the graph of exponential function $e^x.you can use wolfman to draw the graph.

Answer 3

If $a > 0$ then $a^x$ is defined for all $x \in \mathbb R$. (Take by word for it.) The function $a^x$ is continuous (take my word for it) and graphs easily and nicely.

If $a = 0$ then for $x > 0$, $0^x = 0$. This is perfectly acceptable and we can talk and graph $0^x$ on positive $x$ all we want. But this is boring and there is nothing to say as all it is is $0^x = 0$.

If $x =0 $ then $a^x = 0^0$ which is undefined. If $x < 0$ then $0^x = 0^{-|x|} = 1/0^{|x|}$ which is utterly unacceptable.

If $a < 0$ things get bad. If $n$ is odd $a^n < 0$ and $a^{1/n} = \sqrt[n]{a} < 0$ and if $n$ is even $a^n > 0$ and $a^{1/n} = \sqrt[n]{a}$ which does not exist. (Unless we are talking about complex numbers but that is an issue I will not get into here.) If $r = n/m \in \mathbb Q$ with $n/m$ in lowest terms, we will have $a^r$ being positive if $n$ is even and $m$ is odd, being negative if $m$ and $n$ are odd, and undefined if $m$ is even. As every interval $x \in (a)$ will have rationals fitting each of these descriptions, $a^r$ will jump from negative to positive and have undefined spots in every interval no matte how small. So $a^r$ is not always defined and not continuous. For $x$ irrational $a^x$ just can't have any meaning as $x$ irrational is a limit of rational numbers so $a^x$ is a limit of $a^r$s for rational $r$s. But as the $a^r$ jump all over the place and don't always exist. $a^x$ just doesn't mean anything.

So that is why we say $a > 0$.

Now on to the graph.

Two things are important while making a graph: i) the actual value of the point $(x=x, y = a^x)$ of course. and ii) what is the slope of the tangent line as the graph goes through the point. This tells you how steep or shallow to make the graph at that point.

At point $(x, a^x)$ the slope of the tangent is $\ln a*a^x$. Why? Calculus.

Let's consider $1 < a = e \approx 2.718$. Why I chose this value and why choosing this value will be enough to describe all values will hopefully become clear.

$e^x$ has the rather wonderful property that the slope of the tangent line at $(x,e^x)$ is $\ln e * e^x = 1*e^x = e^x$.

So consider, say, $x = -10$ then $e^{-10} \approx 0.00004539992976248485153559151556055$ which is really small and the slope at this point is about $0.00004539992976248485153559151556055$ which darned near flat. By $x = -2$. $e^{-2} \approx 0.13533528323661269189399949497248$ and the slope is $0.13533528323661269189399949497248$ so it's increased a teeny bit but is still mostly flat. By $x = 1$ we have $e^{-1} = 0.36787944117144232159552377016146$ so it's beginning to increase in ernest. At $x = 0$ we get $e^0 =1$ and the slope is $1$ so now graph bowls in a nice rounded shape and starts increase pretty fast. At $x =1$ the point is $(1, 2.718...)$ with a slope of $2.718...$ and by $x=2$ point and slope have reach about $7.3$ and by $x =3$ they are both more than $20$ and the graph explodes... er... exponentially.

So thats the shape; a long flat stretch, a slow increase to a very fast increase.

All graphs of $a^x$ will have this basic shape.

$a^x = e^{\ln a}x$. This stretches or squishes the graph be a factor of $\ln a$. If $a > e$ and $\ln a > 1$ this "shrinks" the graph and makes the graph "smaller and sharper". It stays flat even longer but when it starts to increase it increases faster. At $x = 0$ the point is still $(0,a^0 = 1)$ (that will be true for all $a$) the the slope is $\ln a$ which is steeper than 1. So it blows up quicker.

For $1 < a < e$ then $0 < \ln a < 1$. This stretches the graph out and makes it "gentler". At the point $(0,1)$ (all $a^x$ grow through $(0,1)$) the slope is $\ln a < 1$ so it's increasing more gently. It will hit a point where the slope is one and $x = \ln(1/\ln a)$. But the general affect of this graph is as though you have look at the graph $e^x$ through a magnifying glass.

Okay. If $a = 1$ we have $\ln a = 0$ and $a^x = 1^x = 1$ so the graph is $y = 1$. (Slope is 0 always). This is very boring.

Finally. If $0 < a < 1$. Note: $a^x =(1/a)^{-x}$ and $\ln a < 0$. This "flips" the graph of $1/a = b > 0$ across the y-axis, so that the graph starts at a near infinity and drops to reasonable levels and then flattens to near zero. It does exactly what $b^x$ does but "in the opposite direction".

And ... that's that.

Sorry if this was too informal.

Go to a graphing program and press in the values for $a = 5,4,3,2,1,1/2,1/3,1/4,1/5$. You'll see what I'm talking about. It's a very pretty and informative symmetry.