# Find the intervals on which the function `f(x) = x^2/(3(10 - x))` is increasing and decreasing

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### 1 Answer

A function f(x) is increasing in the interval where the first derivative f'(x) > 0 and it is decreasing where f'(x) < 0.

For the function f(x) = `x^2/(3*(10 - x))`

f'(x) = `(2x*(10 - x) + x^2)/(3*(10-x))`

=> `(20x - 2x^2 + x^2)/(30 - 3x)`

=> `(20x - x^2)/(30 - 3x)`

This is positive when

`x(20 - x)>0` and `30 - 3x > 0`

=> `x(20 - x)>0` and `10 > x`

=> x > 0, 20 > x and 10 > x or x < 0, 20 < x and 10 > x

=> x lies in `(-oo, 10)`

It is also positive when

x(20 - x)<0 and 30 - 3x < 0

=> x(20 - x)<0 and 10 < x

=> x > 0, 20 < x and 10 > x or x < 0, 20 > x and 10 > x

=> x lies in` (-oo, 0)`

The function is decreasing when x lies in `(10, oo)` .

**The function `f(x) = x^2/(3*(10 - x))` is increasing in the interval `(-oo, 10)` and decreasing in the interval `(10, oo)` **