Question

Question #0a15a

Answer 1

• Maximum:
  `(0, 1)`
  `(2pi, 1)`

• Minimum:
  `(pi, -1)`

graph{y = cos(x) [-1.315, 7.335, -1.36, 2.966]}

Explanation:

Finding absolute extrema follows exactly the same process as finding relative extrema. The only extra thing is you'll need to is compare the relative extrema, and see which one is the highest/lowest.

Our first step is to find all critical points (i.e. where `f'(x) = 0`) on our interval. Let's do that:

`f(x) = cos(x)`
`f'(x) = -sin(x)`

`-sin(x) = 0 " when " x = 0, pi, 2pi`.

So our critical values are `0, pi, and 2pi`. These are simple unit circle calculations, so if you didn't get them I'd recommend revisiting that. Try this video:

.

Now, we need to test and see which of these is the biggest/smallest.

Important note here: normally we'd also need to test the endpoints of our interval as well. We didn't need to explicitly do that here since the endpoints were also critical values, but in future problems it's key that you carry out that extra step.

Let's test!

`f(0) = cos(0) = 1`
`f(pi) = cos(pi) = -1`
`f(2pi) = cos(2pi) = 1`

Now, what's your largest value? Well, that's `y = 1`, and so therefore that is your absolute maximum value, which occurs at `x = 0` and `x = 2pi`.

What's your smallest value? That's just going to be `y = -1`, and so that's your absolute minimum value, which occurs at `x = pi`.

In coordinates, your absolute extrema are:

• Maximum:
  `(0, 1)`
  `(2pi, 1)`

• Minimum:
  `(pi, -1)`

Take a look at the graph of `y = cos(x)`. Can you see the absolute maxima and minima?
graph{y = cos(x) [-1.315, 7.335, -1.36, 2.966]}

You may wonder: how can `y = 1` be your absolute maximum if you could get it from two different `x` values? Think back to what an absolute maximum means: it's the highest value your function will achieve on the interval. It doesn't matter how many times it gets there, it only matters how high it gets. Most functions will not achieve the maximum many times over an interval, but `cos(x)` is an oscillating function, and will do that.

Hope that helped :)