How to find MAX number in unseen array using for loop?

numbers[i] refers to the value stored at position i. If you were to use if (i > currentMax) then you would always return the last element, since the last element always has the greatest index.

Don't reinvent the wheel, use Math.max(...numbers).

Say you have an array like:

[1, 2, 4, 2]

This starts by setting currentMax numbers[0] which is one. Then it loops through the array one element at a time. If it finds a larger number during that loop — in other words if (numbers[i] > currentMax) then it sets the currentMax that number instead. For example this will happen the second and thirds times through the loop when i equals 2 & 4. But it won't happen the last time through the loop. An easy way to watch this happen is to print some stuff to the console as it runs:

function max(numbers) {

  let currentMax = numbers[0];
  for (let i = 0; i < numbers.length; i++) {
    console.log("i:", i, "element:", numbers[i], "max:", currentMax)
    if (numbers[i] > currentMax) {
      currentMax = numbers[i];
      console.log("new currentMax:", currentMax)
    }
  }
  return currentMax;
}

max([1, 2, 4, 2])

In this case i is an "index" which allows us to iterate over all the positions in the array (and accessing their values). In this case i=0, i=1,..., i=numbers.length,

if (numbers[i] > currentMax) asks if the number stored in the array in the position i is greater than the currentMax value. This guarantees that from the provided array the maximum number is returned.

If you ask if (i > currentMax) you compare the value of the "index" (i) with the value of the currentMax value. This is incorrect if you want to return the greatest value from an array of numbers.

Like you said, you can reference indexes in arrays by doing numbers. Instead of hard coding numbers you can use a variable that has a number as a value.

function max(numbers) {
  // get the value in the first place in the array  
  let currentMax = numbers[0];

  // create a variable called i
  // set it to 0
  // loop through, increasing i each time, for as long as i is less than the length of the array
  // the first time through i = 0
  // the second time through i = 1
  // then i = 2
  // ... repeat until the end
  for (let i = 0; i < numbers.length; i++) {
    // get the value from the array at the i place
    // if it is greater than the current max
    if (numbers[i] > currentMax) {
      // then set current max to it
      currentMax = numbers[i];
    }
  }

  // return current max
  return currentMax;
}