Question

find the absolute maximum and minimum values of f(x) = x^3 -4x^2 +7x /x on the interval (0,3]

please help on this one?

Answer 1

Well first, do you see any way to simplify the function?

Answer 2

find the diferential?

Answer 3

@jollyjolly0

Answer 4

is this your function? \[f(x) = \frac{ x^3 -4x^2 + 7x}{ x }\]

Answer 5

yes

Answer 6

You should be able simplify that division. Divide everything by x.

Answer 7

x(x^2 -4x+7)

Answer 8

But that entire thing is divided by x. So the x you just factored out cancels.

Answer 9

ok so I am left with x^2-4x+7 ?

Answer 10

sourwing? please help me?

Answer 11

then what u wanna do is take the derivative of that and find where the slope would be zero (horizontal) so that u can see where the maxes and mins are from that interval.

Answer 12

Sorry, I went to eat. Basically, that is a parabola right? Parabolas with a positive second degree term open upwards, so the minimum value is at the vertex. The function increases symmetrically around the vertex, so whichever interval endpoint is furthest from the vertex is the maximum value of the function.

Answer 13

0 technically isn't included in the interval, so watch out.

Answer 14

f(x)=x^2-4x+7
f'(x)=2x-4

Answer 15

2x-4=0

Answer 16

2x=4
x=2

Answer 17

so is that my mimimum value?

Answer 18

since you have only 1 critical point, you have to take second derivative to know what it is
f"(x) = 2 > 0 always ---> concave up in any interval--> x =2 is abs minimum; no abs max; no relative max/min

Answer 19

@Loser66 I forgot to put the multiple choices up.

they gave me

Max: none; min: 4
Max: 7; min: 4
Max: 7; min: 3
Max: none; min: 3
None of these

Answer 20

so i should choose none of these?

Answer 21

@phi

Answer 22

They want
the absolute maximum and minimum values of f(x) = x^3 -4x^2 +7x /x on the interval (0,3]

which simplifies to f(x)= x^2 -4x + 7 (x≠0)
you found a min at x=2, where f(2)= 3

you should also check the value of f(x) at the boundaries to find the max value *in this interval*

Answer 23

right so that answwer isn't there

Answer 24

there is no answer with min: 2

Answer 25

so i choose "none of these" right?

Answer 26

the minimum value of f(x) occurs at x=2. the minimum value itself is 3

Answer 27

so I need to find the max and its either 7 or 0

Answer 28

if f(0) is undefined that means it has to be 7 right?

Answer 29

You seem to be confusing the x value with the value of f(x)
the max or min is f(x)

f(0) is undefined, but we can use the limit of f(x) as x->0
In practice, use x=0 in the equation

so you want to evaluate f(0) and f(3)
one or the other will be the maximum value.

A picture is worth a thousand words... but calculus gives you the same answer.

Answer 30

I have a lot of trouble with the calculus part but the graph was perfect. thank you

Answer 31

I went and checked, and technically this function does NOT have a maximum.
The idea is that f(x) never reaches the value of 7... there are an infinite number of values that get closer and closer to 7, but never reach it.

so the correct answer is no maximum, a minimum of 3

Answer 32

wow ok. how did you check?

Answer 33

I went googling to double check.... I half-remembered that the open interval caused some kind of problem...
I found this
http://answers.yahoo.com/question/index?qid=20110325091909AA0SPcF

Answer 34

thank you. thank you sooo much

Answer 35

@phi could you give me a quick reminder on how to find the concavity of a curve at a point
?

Answer 36

I remember the test by remembering that
y= x^2 is a "smile" and has a minimum
y' = 2x gives the slope at any point x
y'' = 2 (2nd derivative) is positive. so y''>0 means MIN

y'' <0 means MAX

y'' = 0 might mean INFLECTION POINT

Answer 37

or see this
http://www.math.hmc.edu/calculus/tutorials/secondderiv/

Answer 38

Or see Khan's videos (they tend to be short)
Here is the first in the section on critical points.
http://www.khanacademy.org/math/calculus/derivative_applications/critical_points_graphing/v/minima--maxima-and-critical-points

I would watch that one and maybe a few of the others.

Answer 39

ok thank you very very much