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This is quite doable by hand. As usual, you want to find the derivative and determine where it is 0. The derivative is a cubic polynomial, but thankfully it has an obvious root at [math]x=4[/math], so you're left with a simple quadratic equation to solve.[math]p(x)=x(x-1)(x-4)^2[/math]Don't expand this into a sum of powers of [math]x[/math][math][/math]. Leave it as a product and use the product rule.[math]p'(x)=(x-1)(x-4)^2+x(x-4)^2+2x(x-1)(x-4)[/math]So you see there's a factor of [math](x-4)[/math] everywhere, meaning the derivative is [math]0[/math] at [math]x=4[/math]. Dividing through by that factor, you’re left with a quadratic:[math]q(x)=(x-1)(x-4)+x(x-4)+2x(x-1)=4x^2-11x+4[/math]The roots of [math]q(x)=0[/math] are [math](11\pm \sqrt{57})/8[/math], both of which are less than [math]4[/math][math][/math]. There are three distinct critical points, and the middle one (the larger root of [math]q[/math]) is the maximum.(The original polynomial is of even degree with positive leading coefficient, so it must have two minima and a maximum between them).

Yes, there’s only one solution: [math]x=6[/math]. The other ‘solution’, [math]x=3[/math], which arises after squaring both sides, is only valid for the equation [math]x-2=(x-4)^2[/math], which is not equivalent to the equation [math]\sqrt{x-2}=x-4[/math].Squaring both sides of an equation is a tricky operation. For example, suppose I tell you that [math]a=2[/math]. How many solutions does this equation have? Well, clearly it has only one solution. But now suppose I square both of its sides, to get [math]a^2=4[/math]. How many solutions does this equation have now? It has two solutions: [math]a=2[/math] and [math]a=-2[/math].Thus, squaring both sides may introduce new solutions that weren’t valid for the original equation. This is exactly what happened when we squared both sides of the equation [math]\sqrt{x-2}=x-4[/math] to obtain [math]x-2=(x-4)^2[/math]: the original equation had only one solution, [math]x=6[/math], while the second has two solutions, [math]x=6[/math] and [math]x=3[/math]. This is something that needs to be kept in mind when squaring both sides of an equation.

I wrote it many, many years ago. I was teaching at a few art schools throughout the greater Sacramento area, and I used it to engage my students in the topic of graphing. One of my coolest students (Mr. Wilkinson aka i_luv_ur_mom) posted it to Reddit back in 2011 and it went viral. I'm a full time professor over at American River College, doing every thing I can to make math as enjoyable as possible.***Per a Request, the following comment has been added to the answer on 8/22/15***"If you want a nicer Batman Equation, here's the one I use to teach graphing these days. It's a little cleaner, using the niceness of a radius 5 circle stretched into an ellipse. Be aware that most implicit graphers will require you to break it up using the zero product property, as they really don't like graphing the product of imaginary numbers with zero. Enjoy!"Raw Equation: [math](3 * sqrt(4 - (abs(x) - 2)^2 [/math][math][/math][math]) [/math][math][/math][math]+ abs(x) - 20 - 4*y) * (3 * sqrt(4 - (abs(x) - 6)^2 [/math][math][/math][math]) [/math][math][/math][math]+ abs(x) - 20 - 4*y) * (x^2 [/math][math][/math][math]+ 4*y^2 - 100 * sqrt(abs((7 - abs(2*y - 1))) [/math][math][/math][math]/ (7 - abs(2*y - 1) [/math][math][/math][math]))) * ([/math][math]2[/math][math] * (abs(x) - 3)^2 - 9 * y [/math][math][/math][math]+ 18 * sqrt(abs((2 - abs((abs(x) - [/math][math]4[/math][math])))) [/math][math][/math][math]/ ([/math][math]2[/math][math] - abs((abs(x) - [/math][math]4[/math][math])) [/math][math][/math][math]))) * (-68 * abs((abs(x) - 3/2)) - 9*y [/math][math][/math][math]+ 54 * sqrt(abs((43 - abs((136 * abs(x) - 229)))) [/math][math][/math][math]/ (43 - abs((136 * abs(x) - 229)) [/math][math][/math][math]))) * (y - 5 * sqrt(abs((1 - abs(x))) [/math][math][/math][math]/ (1 - abs(x) [/math][math][/math][math]))) = 0[/math].