Graphing the function $(-2)^x$

Question

When I wanted to graph $y=(-2)^x$ many graphing calculator apps refused to plot it. TI-Nspire CAS plotted it as shown in the first picture. I think the plot is not correct as only the envelopes should be there with no values between the envelopes as shown in the second picture and the $(-2)^x$ should look like $2^x$ and $-(2^x)$ plotted on the same graph but of course with many discontinuities as explained in my analysis below. Am I right?

$(-2)^x$ as graphed by the TI-Nspire

this is what I think it should be

Here’s my analysis of the function:

f(x)=(-2)^x

First: when x >=0

• if x is an integer, x>=0

(-2)^0=1

(-2)^1=-2

(-2)^2=4

(-2)^3=-8

(-2)^4=16

(-2)^x oscillates back and forth When x is an even integer (-2)^x is positive. When x is an odd integer (-2)^x is negative.

• when x is a rational number, x>0 let x=p/q , p>0, q>0

(-2)^x=(-2)^(p/q)=((-2)^p)^(1/q)

if p is even and q is odd, (-2)^x is a positive real value Example: (-2)^(100/51)= 3.8927

if p is odd and q is odd, (-2)^x is a negative real value. Example: (-2)^(99/51)=-3.8402

if p is odd and q is even, (-2)^x is an imaginary value (not defined in the set of real numbers) Example: (-2)^(99/50)=i 3.9449

(Also in all the above cases, if you extend your analysis to include complex, we get q complex roots.)

In the domain of rational numbers, (-2)^x oscillate or is undefined (imaginary or complex)

• When x is irrational, x>0 (-2)^x has no real value. It has an infinite number of complex roots.

Second: when x<0 (-2)^x=1/(-2)^|x| Use the same approach above to analyze the behavior of the function.

Answer 1

The graph of $f(x)=(-2)^x$ is problematic for real numbers $x$. Think about what happens when $x=\frac{1}{2}$. Then $f(x)=(-2)^{\frac 1 2}=\sqrt{-2}$. Can you see why this is a problem to graph?

The graph of $f(x)=(-2)^x$ only makes sense for integer values of $x$. Also, as zwim pointed out in the comments, your second graph is not the graph of a function, as it is multivalued, i.e. one input of $x$ gives two outputs of $f(x)$.

Answer 2

$(-1)^x$ as defined in the real domain is only making sense for $x\in\mathbb Q_\text{odd}$

$=\{\frac pq\mid p\in\mathbb Z,q\in\mathbb N^*,\gcd(p,q)=1,\text{ and }q\text{ is odd }\}$

Have a look at this post for instance : Exponential Equation with base -1, 0, and 1

Thus the graph of $(-2)^x$ has holes on all points of ${\mathbb Q_\text{odd}}^\complement$.

More than a continuous graph with infinity of holes it is in reality more of a collection of isolated points in the plane. But I agree that thet are so close from each other that visually it looks like the curve you represented. But if you could zoom up with infinite factor you would see a dashed line.