I am not able to
Plot[(-1)^x + 2^x - 2 x - 1, {x, -4, 4}]
directly. Some rework might be needed with complex branch cuts perhaps.
There are real roots at $ x=3, x=2 $. I configured this expression with the first root 3 and there are several other concomitant complex roots. It comes to me as a surprise so with the infallible Mathematica. What may be something that I miss? I wish to be able to see the plot and all roots in the domain with your help.
Here you are plotting a complex function, as you know, so you need to plot the real and imaginary parts separately.
Plot[ReIm[(-1)^x + 2^x - 2 x - 1], {x, -4, 4}, Evaluated -> True]
You can get some help on this from Wolfram Alpha, if you start your expression with two equal signs:
This shows that you need to plot the real and imaginary parts separately.
When in doubt check for Complexes,
FunctionDomain[(-1)^x + 2^x - 2 x - 1, x, Complexes]
True
FunctionRange[(-1)^x + 2^x - 2 x - 1, x, Complexes]
-2. <= Complexes <= 4.25
ComplexExpand[(-1)^x + 2^x - 2 x - 1, x]
$e^{-\pi \Im(x)} \cos (\pi \Re(x))+i \left(e^{-\pi \Im(x)} \sin > (\pi \Re(x))+2^{\Re(x)} \sin (\log (2) \Im(x))-2 > \Im(x)\right)+2^{\Re(x)} \cos (\log (2) \Im(x))-2 \Re(x)-1$
Find Re and Im,
sol1 = ComplexExpand[Re[(-1)^x + 2^x - 2 x - 1]]
$-2 x+2^x+\cos (\pi x)-1$
sol2 = ComplexExpand[Im[(-1)^x + 2^x - 2 x - 1]]
$\sin (\pi x)$
Have a plot:
Plot[sol1 - sol2, {x, -5, 5}]
And solve for roots:
eqn = sol1 - sol2
$-2 x+2^x-\sin (\pi x)+\cos (\pi x)-1$
soln = NSolve[{eqn, 2 <= x <= 4}, x, Reals]
{{x -> 2.}, {x -> 3.}}
And you can plot the Functions:
Plot[{sol1, sol2}, {x, -4, 4}]
Plot[eqn, {x, -4, 4}]
Nice Plot if you need to viz other roots:
Plot[{eqn, -eqn}, {x, -4, 4}]
NSolve[(-1)^x + 2^x - 2 x - 1 == 0 && -4 < Re@x < 4 && -1 < Im@x < 1, x]– Michael E2 May 04 '16 at 17:27