Wednesday

September 17, 2014

September 17, 2014

Posted by **Billy Baston** on Monday, March 25, 2013 at 7:49pm.

- absolute minimum math -
**Reiny**, Monday, March 25, 2013 at 8:26pmy = x^2/(x-3)

dy/dx = ( (x-3)(2x) - x^2) /(x-3)^2

= (2x^2 - 6x - x^2)/(x+3)^2

= (x^2 - 6x)/(x-3)^2

for a max/min, dy/dx = 0

x^2 - 6x = 0

x(x-6) = 0

x = 0 or x = 6

so in the interval given, we need x = 6

if x = 6

y = 36/3 = 12

The minimum is 12

**Answer this Question**

**Related Questions**

Calculus (pleas help!!!) - Find the absolute maximum and absolute minimum ...

Calculus (pleas help!!!) - Find the absolute maximum and absolute minimum values...

Calculus - Find the absolute maximum and absolute minimum values of the function...

Calculus - Find the absolute extrema of the function on the interval [2, 9]. (...

Calculus - Determine where the absolute extrema of f(x)=-3(x^2)+7x on the ...

Calculus - Find the absolute maximum and absolute minimum values of f on the ...

Calculus - Find the absolute minimum and absolute maximum values of f on the ...

Calculus - Find any absolute maximum and minimum and local maximum and minimum ...

Calculus - Find the absolute maximum and absolute minimum values of f(x)=sinx...

Calculus - Find any absolute max/min and local max/min for the function f(x)=x^3...