Question

What is the minimum value of `g(x) = x^2-2x - 11/x?` on the interval `[1,7]`?

Answer 1

The function is continuously increasing in the interval `[1,7]` its minimum value is at `x=1`.

Explanation:

It is obvious that `x^2-2x-11/x` is not defined at `x=0`, however it is defined in the interval `[1,7]`.

Now derivative of `x^2-2x-11/x` is `2x-2-(-11/x^2)` or

`2x-2+11/x^2` and it is positive throughout `[1,7]`

Hence, the function is continuously increasing in the interval `[1,7]` and as such minimum value of `x^2-2x-11/x` in the interval `[1,7]` is at `x=1`.

graph{x^2-2x-11/x [-40, 40, -20, 20]}