00:01
This problem wants us to find the absolute and local maximum and minimum values of f.
00:04
And when we look at our function f, we're given that f of x equals 2 plus x minus 1 over x.
00:10
And we're limited to the domain of 1 less than or equal to x less than or equal to 4.
00:14
And to find our absolute minimum and maximum value, what we're going to do first is take the derivative of this function.
00:20
And we're going to take the derivative by first taking the derivative of our constant 2.
00:24
And the derivative of any constant is just 0, so 2 is gone.
00:28
And then we have plus x that has an understood 1 in front.
00:31
And that's a linear term, and the derivative of a linear term is just the coefficient, so that's 1.
00:37
And then for minus 1 over x, we're going to bring x to the top of our fraction as x to the negative first.
00:42
And then we'll take our negative 1, multiply by our negative in front, which is going to give us positive understood 1.
00:49
And then times x to the negative first minus 1 for the exponent, which is x to the negative second.
00:54
So we're left with a derivative expression of 1 plus 1 over x squared.
01:02
And the reason the derivative is going to help us with maximum and minimum is because when we set our derivative to 0 and solve for the x values that make that true, that gives us the possible locations of minimums and maximums for our function...