How can I caculate the range of value of a function by matlab?

I want to find the interval of values in a symbolic function. Suppose that there is a function f(x,y,z)=x^z+2*y^3. And the x>1, 4>y>-3, 2>=z>=-1. How can I use matlab to get the f(x,y,z) range of value? Or could you tell me which code I can get my target? Thank you.

I just reach the step as follow:

syms x y z

f(x,y,z)=x^z+2*y^3

Answer 1

[Walter Roberson]

You need to use calculus to find all the critcal points -- the places where the derivative are 0. Figure out which of the critical points are within the boundaries. Evaluate the function at each of those in-range critical points. Now, also evaluate the function at each combination of boundaries -- for example, minimum x, minimum y, maximum z; minimum x, maximum y, minimum z, and so on. You will end up with 8 boundary values, together with however-many critical points. Now you can take max() and min() to evaluate the range of the function.

syms x y z
assume(x > 1);
assume(-3 < y & y < 4)
assume(-1 <= z & z <= 2)
f = x^z+2*y^3;
dx = diff(f,x)

dx = 

xcrit = solve(dx, x, 'returnconditions', true)

xcrit = struct with fields:
             x: [0×1 sym]
    parameters: [1×0 sym]
    conditions: [0×1 sym]

dy = diff(f,y)

dy = 

ycrit = solve(dy, y, 'returnconditions', true);
[ycrit.y,  ycrit.conditions]

ans = 

dz = diff(f,z)

dz = 

zcrit = solve(dz, z, 'returnconditions', true)

zcrit = struct with fields:
             z: [0×1 sym]
    parameters: [1×0 sym]
    conditions: [0×1 sym]

x would normally have a critical point at x = 0, since 0^positive = 0, but the assumption is that x > 1

x does have a critcal point at x = infinity and z < 0, which solve() has missed.

A moment thought experiment: since x can be infinite, and z can be positive, then the upper range of the function has to be infinity. The x^z term cannot be negative for positive x, but infinity^-1 is 0 so it can be 0; therefore the minima of the function is determined by the 2*y^3 term, and since y can be negative, the minima of that is going to be at the most-negative y.

The dy solutions of y = 0 are inflection points.

[Ziwei Li]

Thank you ! I understand your solution. I am sorry for the exmple is a bit weird.