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We will take the first derivative and equate the result to zero, to identify the critical point/s.
`f ' (x) = 1 - 1/x`
`f'(x) = (x - 1)/x` (Taking note that the lcd is x).
`x - 1 = 0`
Add 1 on both sides.
`x = 1` So, we have a critical point at x = 1.
Take note that we will also have a critical point on the value of x where our functon is undefined. So, we equate the bottom to zero as well.
x = 0.
But, x = 0, is not on our interval since, we have[1/2, 2].
So, our only critical point is at x = 1.
Now, we will plug-in the endpoints of our interval and our critical points on our original function.
`f(1/2) = (1/2) - ln(1/2) = 1.193147181 `
`f(1) = (1) - ln(1) = 1`
`f(2) = 2 - ln(2) = 1.306852819`
So, we will have an absolute minimum at x = 1, and its y = 1. Our absolute maximum will be at x = 2, which is y = 1.3069.
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