Hello, I'd like to define this function. Any help please?

I tried making use of the if function:

function u1_problem(x)
    x=1;
    
    if x >= 1 && x <= 1.5
        u1=@(x) 3.*(x-1);
    end
    
    if x > 1.5 && x <= 2
    u1=@(x) 3-x;
    end
    
    if x > 2
        return
    end
    x=x+1;
    
end

[David Hill]

Is your function only defined from 1 to 2? Your function does not make any sense when x = 1.5.

[madhan ravi]

Why not https://www.mathworks.com/help/symbolic/piecewise.html ?

Answer 1

[Steven Lord]

A couple comments:

function u1_problem(x)

Your function doesn't return any values to its caller. So whatever variables you define inside the function are discarded when the function finishes its execution.

    x=1;

This overrides whatever the user who called your function passed into your function as input.

    if x >= 1 && x <= 1.5

If x must be a scalar, this would work. But if it can be a non-scalar (usually a vector or matrix) it's not going to work.

        u1=@(x) 3.*(x-1);
    end

This makes u1 a function handle. Do you want to return a function handle or do you want to return a value? [This assumes you modify your code to return u1.]

    
    if x > 1.5 && x <= 2
    u1=@(x) 3-x;
    end
    
    if x > 2
        return

So your function isn't going to define u1 in the case where x is greater than 0 or if x is not greater than or equal to 1?

    end
    x=x+1;

This modifies the value of x inside this workspace then the function immediately exits and discards the copy of x in this workspace.

    
end

Answer 2

[William Alberg]

Im not entirely sure if this is what you are looking for.

syms x
u = @(x) func(x)


function y = func(x)
    
    if 1 <= x && x <= 1.5
        y = 3*(x-1);
    elseif 1.5 < x && x <= 2
        y = 3-x;
    else
        y = nan;
    end
end

You can also just use func(x) direcly