Question

Find the minimum algebraically

Answer 1

@zepdrix

Answer 2

So to find max/min within an interval we need to first find `critical points`.
Then we'll plug our `critical points` and `boundary end points` back into the original function and compare the outputs.

The largest output will be our maximum, while the smallest will be our minimum.

Answer 3

To find critical points, we need the derivative of f(x).
Have you tried taking its derivative? :)

Answer 4

yes i got (1-(125)*(x+5)-(125x)(x))/(x+5)^2

Answer 5

when i set it equal to 0, there is no solution

Answer 6

@zepdrix

Answer 7

\[\Large f(x)\quad=\quad x-\frac{125x}{x+5}\]Taking the derivative, applying quotient rule to the second term, gives us:\[\Large f'(x)\quad=\quad 1-\frac{125(x+5)-125x(1)}{(x+5)^2}\]

Answer 8

Derivative of (x+5) should give you (1) in that second upper half.

Answer 9

yup thats the derivative i got

Answer 10

Then it looks like we have some nice simplifications from there.

Answer 11

\[\Large f'(x)\quad=\quad 1-\frac{625}{(x+5)^2}\]

Answer 12

yes I got that...and when i tried to find the minimum i got something like -72.62..

Answer 13

Setting our derivative equal to zero, and finding a common denominator:\[\Large 0\quad=\quad \frac{(x+5)^2-625}{(x+5)^2}\]

Answer 14

there is no solution right?

Answer 15

Normally we would want to include points of discontinuity that the derivative contains.
But in this case, the point x=-5 isn't included in the domain of the original function, so we'll only worry about the numerator here.

\[\Large 0\quad=\quad (x+5)^2-625\]Ok I guess we need to expand out the square.

Answer 16

wait are'nt we supposed to cross multiply?

Answer 17

Oh wait there's a much easier way than expanding out the square :P hah.
Cross multiply? From where? From here?\[\Large 0\quad=\quad \frac{(x+5)^2-625}{(x+5)^2}\]

Answer 18

yeah

Answer 19

Sure you can cross-multiply from there if you like :)

Answer 20

when you do you get 0

Answer 21

No no no. Careful!

\[\Large 0\ne \frac{0}{0}\]
\[\Large 0=\frac{0}{1}\]We can use this fraction for cross-multiplication.

Answer 22

0*1=0

Answer 23

\[\Large \frac{0}{1}\quad=\quad \frac{(x+5)^2-625}{(x+5)^2}\]Cross-multiplying gives us:\[\Large 0(x+5)^2\quad=\quad 1\left[(x+5)^2-625\right]\]Right?

Answer 24

right

Answer 25

wait why is the x+5 in the top?

Answer 26

wait nvm

Answer 27

sorry

Answer 28

i understand why

Answer 29

Ok :)
So I guess that leads us to:\[\Large 0\quad=\quad (x+5)^2-625\]

Answer 30

+625

Answer 31

actually just leave it

Answer 32

Mmm I think it's -.. lemme make sure I didn't miss a sign somewhere.

Answer 33

you cane combine terms

Answer 34

oh no +625 meant add it to the other side but i realized you can just expand the binomial and then combine the terms lol

Answer 35

Oh I see what you meant :)
Yes adding 625 to each side is the best approach.

Answer 36

really? okay

Answer 37

\[\Large 625\quad=\quad (x+5)^2\]

Answer 38

x^2+10x+25

Answer 39

idk if its correct... did it in my head

Answer 40

Let's not expand out the binomial.
After we move the 625, let's take the square root of each side.

Answer 41

oh okay so its 25

Answer 42

x+5=25

Answer 43

x=20

Answer 44

Ok good!\[\Large \pm25\quad=\quad x+5\]

Answer 45

or c=-20

Answer 46

x=-20

Answer 47

oh wait....i did something wrong

Answer 48

We took the root of a `squared` term so we need to include the +/-.

So ummm....\[\Large x=\pm25-5\]

Answer 49

right....sorry i have no paper with me to write it out

Answer 50

LOL :) Get some paper ya bum! lolol

Answer 51

lollll okay!

Answer 52

So we'll use +25 and then -25 to find our 2 critical points for x .

Answer 53

alright so do we plug this value into the original function?

Answer 54

isnt it +25-5 and -25-5?

Answer 55

so wouldnt the critical pointd be 20 and -30?

Answer 56

Yes very good!
Do both of those x values fall in our given interval, \(\Large x\in[0,\;23]\) ?

Answer 57

well only 20

Answer 58

Ok so we'll throw away the x=-30 point, we can't use that.

Answer 59

right

Answer 60

So now we can go about finding out which point is a max, min or whatever else.
We have 3 values to compare, the `end points` and our `critical point`.
x=0, x=20, x=23

Answer 61

\[\Large f(0)=?\]\[\Large f(20)=?\]\[\Large f(23)=?\]

Answer 62

f(20)=-80?

Answer 63

f(0)=0

Answer 64

f(23)=-336.375

Answer 65

so would f(23) be the min?

Answer 66

Hmm you didn't calculate f(23) correctly.
The other two points look good though.

Answer 67

\[\Large f(0)=0\\ \Large f(20)=-80\\ \Large f(23)=?\]

Answer 68

-79.67?

Answer 69

Good good good.

Answer 70

You are a lifesaver! Thank you. You should be a moderator or something lol

Answer 71

Here is a fun little graph of the points in case you wanted to see.
https://www.desmos.com/calculator/utxipllxei

Answer 72

Yay, glad to help \c:/

Answer 73

oh wow that website looks helpful..thanks for that as well :)