## Question:

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

f(x) = x / x^2 − x + 4 [0, 6]

absolute minimum value

absolute maximum value

?

f(x) = x / x^2 − x + 4 [0, 6]

absolute minimum value

absolute maximum value

?

### Tag:

**Asked About:**Stewart Calculus: Early Transcendentals, Pg 281, Ex 54

I am presuming that you intended to write: f(x)=x/(x^2-x+4)

First find the derivative of f(x). Then set it to zero. When you multiply the denominator by zero, it gets zeroed out, so we can ignore it. In other words, the only time the derivative will be zero, is when the numerator is equal to zero. So set that to be zero.

You should get -x^2+4=0. That means that x^2=4; x= +2 and -2. Since the Interval is I[0,6] negative 2 gets thrown out. Now we test the endpoints and critical points of the function.

f(0)=0

f(2)=1/3 (0.33333)

f(6)=3/17 (0.176)

So the maximum is1/3, which is achieved at x=6

and the minimum is 0, which is achieved at x=0

First find the derivative of f(x). Then set it to zero. When you multiply the denominator by zero, it gets zeroed out, so we can ignore it. In other words, the only time the derivative will be zero, is when the numerator is equal to zero. So set that to be zero.

You should get -x^2+4=0. That means that x^2=4; x= +2 and -2. Since the Interval is I[0,6] negative 2 gets thrown out. Now we test the endpoints and critical points of the function.

f(0)=0

f(2)=1/3 (0.33333)

f(6)=3/17 (0.176)

So the maximum is1/3, which is achieved at x=6

and the minimum is 0, which is achieved at x=0

Matt Bausman

5yrs
I should also note that there may be a Critical Point where the derivative does not exist (ie the root of a negative or dividing by zero), but that does not appear to happen in this case