How to visualize a function with four arguments?

If I have a function, f[a, b, c, d], with a, b, c and d varying between 0 and 1 continuously. What would be the best way to explore this function visually with Mathematica? Is evaluating multiple Plot3D expressions with two of the four arguments held constant my only option?

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For example:

Manipulate[
ContourPlot3D[Norm[{x, y, z}]^ (3 + w),
{x, 0, 1}, {y, 0, 1}, {z, 0, 1},
ContourStyle -> (Directive[Opacity[.3, #]] & /@ {Red, Green, Cyan}),
Contours -> {1, 2, 3}, MeshStyle -> None],
{w, 0, 1}]


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Nice!!! Further examples plus mathematical proofs can be found in the last 3 chapters of link –  eldo May 28 at 18:19
I'm having difficulties interpreting the contour surfaces in this plot. What do they represent exactly? How can I get a feeling of how my function behaves from the contour surfaces? –  su1 Jun 2 at 14:34
@su1 "get a feeling" is quite vague :) What exactly do you want to explore? –  belisarius Jun 2 at 14:45
for example understand from the contour surfaces how my output varies when x increases, knowing all other inputs are fixed. Basically I don't really know how to read a ContourPlot3D. It's easier to read it when it's in the form ContourPlot3D[f==g, ...] but when in the form of ContourPlot3D[f, ...] I don't know what the plot represents –  su1 Jun 2 at 15:16
@su1 In the form I used the function you've three contours: On the red one the function value is 1, on the green one the value of the function is 2 and in the cyan contour the function evaluates to 3.Is it clear now? –  belisarius Jun 2 at 15:20

Manipulate is a great tool anyway:

Manipulate[
ContourPlot3D[
c - Sin[d] - a x^2 - b y^2 + z^2 == x y z, {x, -2, 2}, {y, -2,
2}, {z, -2, 2}, Mesh -> None,
ColorFunction -> Function[{x, y, z}, Hue[z]],
ContourStyle -> Opacity[0.75]], {a, 0, 1}, {b, 0, 1}, {c, 0, 1}, {d,
0, 1}]


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If I'm not mistaken, what you're visualizing there is a function of seven arguments. –  Rahul May 28 at 22:58