Find the Absolute Maximum and minimum values of f on the given interval.

f(x) = x - 2(tan^-1)x on interval [0,4]

### 1 Answer | Add Yours

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.**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes