Find the absolute maximum and absolute minimum values of *f* on the given interval.

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You need to remember Fermat's theorem that tells you about the relation between the critical points and extrema of a function.

You may find the critical points of a function solving the equation `f'(x)=0` , hence you need to differentiate the function with respect to x to find the equation of `f'(x)` such that:

`f'(x) = 1 - 9/(x^2)`

You need to solve the equation `f'(x)=0 ` such that:

`f'(x)=0 =gt 1 - 9/(x^2) = 0`

You should identify the difference of squares such that:

`1 - 9/(x^2) = (1 - 3/x)(1 + 3/x) = 0`

`1 - 3/x = 0 =gt x - 3 = 0 =gt x = 3`

`1 + 3/x = 0 =gt x = -3`

Notice that the value -3 is not in interval (0.2 , 12), hence you need to evaluate the value of function at critical value x=3 such that:

`f(3) = 3 + 9/3 = 6`

**Since the values of derivative are negative between `(-3,3)` and positive for x smaller than -3 and larger than 3, then the functiop decreases in `(0.2 , 3)` and increases in `(3 , 12` ), hence, the function reaches its absolute minimum at x = 3.**

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