How to find a value of a function in MATLAB?

Question

I just started using MATLAB, and for that I am not familiar with MATLAB itself. My sample code is as follows:

function Problem1 = BisectionMethod1()
    a = input('enter function:', 's');
    f = inline(a);
    iteration_counter = 0;
    al = input('enter left bound: ');
    ar = input('enter right bound: ');
    break;
    disp(f('al'))
    disp(f('ar'))

When I set a as x+1, and set my left and right bound as 1 and 2, it displays f(al) and f(ar) correctly.

The problem seems to begin when I have a coefficient in front of 'x'.

For example, when I set a as 2x+1, and set my left and right bounds as 1 and 2, MATLAB would give me error.

Like I said, I am new to MATLAB, is there any way to solve this?

Answer 1

One point I need to make is that multiplication requires a * operator. By doing 2x , MATLAB would interpret this as a variable named 2x and MATLAB does not support variables where there is a number that comes first. Therefore, you need to do 2*x + 1 . In addition, you need to remove your break statement. Your code will exit prematurely if you leave this in.

Also, simply remove the single quotes when calling f . You are inputting the variable, not the actual name of the variable itself. As such, you would do:

disp(f(al));
disp(f(ar));

Using your code, this is what I get:

>> a = input('enter function:', 's');
enter function:2*x + 1
>> f = inline(a);
>> al = input('enter left bound:');
enter left bound:1
>> ar = input('enter right bound:');
enter right bound:2;
>> disp(f(al))
     3

>> disp(f(ar))
     5