Hello mic34!
The function 3(x-2)^2/(x+2)(x-3) has no absolute extrema en in the
interval (-2,3), because it tends to infinity as x approaches either
-2 or 3 (note that you are dividing by numbers close to 0 as x
approaches -2 or 3).
In order to find relative extrema, we have to take the derivative of
the function and equate it to 0. So,
f(x) = 3(x-2)^2/((x+2)(x-3))
f'(x) = 3[2(x-2)(x+2)(x-3)-(x-2)^2(x-3)-(x-2)^2(x+2)]
---------------------------------------------
[(x+2)(x-3)]^2
Taking x-2 as common factor in the numerator, we get
3(x-2) [2(x+2)(x-3)-(x-2)(x-3)-(x-2)(x+2)]
------------------------------------------
[(x+2)(x-3)]^2
Finally, distributing what is inside brackets in the numerator and
cancelling out terms, we get that
f'(x) = 3(x-2)(3x-14)
-------------
[(x+2)(x-3)]^2
Now, as long as x is not either -2 or 3, the denominator is a positive
number. So, in order to find and extrema we have to find where is the
numerator equal to 0. Clearly, it's 0 whenever x=2 or whenever 3x=14
(x=14/3). We don't care about the 14/3 because it's larger than 3. So,
there is a relative extrema in x=2.
Is it a maximum or a minimum? Well, as we have seen the denominator is
positive. The term (3x-14) is negative 'near' x=2. So we have to check
what happens with the derivative to the left and to the right of x=2.
As x approaches 2 from the left, the term (x-2) is negative, so the
whole derivative is positive, thus the function is increasing as x
approches 2 from the left. As x approches 2 from the right, we have
that (x-2) is positive, so the whole derivative is negative, thus the
function is decreasing to the right of x=2. Therefore, x=2 is a
relative maximum. (Alternatively, you can find f''(2) and check wether
it's positive or negative to see if it's a minimum or a maximum)
Best regards,
elmarto |
