Find the Absolute Maximum and minimum values of f on the given interval.
f(x) = x - 2(tan^-1)x on interval [0,4]
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Take the derivative of the given function, and equate it to zero.
`f '(x) = 1 - 1/(1 + x^2)` Take note that derivative of tan^-1(x) = dx/(1 + x^2).
`1 - 1/(1 + x^2) = 0`
Multiply both sides by 1 + x^2.
`1 + x^2 - 1 = 0`
Combine like terms.
`x^2 = 0`
Take the square root of both sides.
`x = 0`
Plug-in the x = 0, and x = 4 on the original fucntion.
`f(0) = (0) - tan^-1(0) = 0`
`f(4) = 4 - tan^-1(4) = 5.33 ` radians
So, we will absolute min at x = 0, and the value is zero.
And we will have absolte max at x = 4, and the value is 5.33 radians.
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