Question

Calculus problem: find the values of constants a, b, and c so that the graph y=(x^2+a)/(bx+c) has a local max at x=3 and a local min at (-1,2)

Answer 1

Have you differentiated y yet?

Answer 2

So far, I have...y=(9+a)/(3b+c) by subing in x=3
2b-2c-a=1 by subing in (-1,-2) and subbing in x=3
b=ab+2c from the first derivative and subbing in (-1,-2)

Answer 3

2b-2c-a=1 by subing in (-1,-2) **

Answer 4

c=(ab-9b)/6 from the first derivative and subbing in x=3

Answer 5

we know we will have f'(x)=0 when x=3
and f'(x)=0 when x=-1
You are also given (-1,2) is on the graph of y=f(x)
Also I'm slightly confused it the point (-1,2) or (-1,-2)?

Answer 6

oops im sorry the point is (-1,-2)

Answer 7

what did you get for y'?

Answer 8

and yeah, i used that info and got equations... but the problem is i have to many unknowns so if i substitute the equations in, ill still have unknowns lol. y'=[x(bx+2c)-ab]
/(bx+c)^2

Answer 9

you will have 3 equations with 3 variables

Answer 10

\[y'=\frac{(x^2+a)'(bx+c)-(bx+c)'(x^2+a)}{(bx+c)^2} =\frac{2x(bx+c)-b(x^2+a)}{(bx+c)^2} \\ y'=\frac{2bx^2+2cx-bx^2-ba}{(bx+c)^2}=\frac{bx^2+2cx-ba}{(bx+c)^2}\]
your y' looks wonderful :)

Answer 11

so we will have one equation given be f(-1)=-2
and two equations given by f'(x)=0 when x=3 then also at x=-1

Answer 12

\[-2=\frac{1+a}{-b+c} \text{ from the } f(-1)=-2 \text{ thing } \\ 0=\frac{9b+6c-ba}{(3b+c)^2} \text{ from the } f'(3)=0 \\ 0=\frac{b-2c-ba}{(-b+c)^2} \text{ from the } f'(-1)=0\]

Answer 13

those second two equations we can't assume the bottom doesn't equal in zero and rewrite the equations

Answer 14

and let's get rid of the fraction for the first one too

Answer 15

\[2b-2c=1+a \\ 0=9b+6c-ba \\ 0=b-2c-ba\]

Answer 16

did you get this far?

Answer 17

yup!

Answer 18

\[0=-2b+2c+1+a \\ 0=9b+6c-ba \\ 0=b-2c-ba \]
Well right away I can see how we can get a system of two equations with two variables

Answer 19

\[0=9b+6c-ba \\ 0=b-2c-ba \\ \text{ multiply bottom equation by } -1 \\ 0=9b+6c-ba \\ 0=-b+2c+ba \\ \text{ add equations together } \\ 0=8b+4c \\ 8b=-4c \\ c=-2b \\ \text{ and the first equation was } 0=-2b+2c+1+a \\ \text{ so we can sub in } c=-2b \text{ in that one } \\ 0=-2b+2(-2b)+1+a\]
and lied about the system of two equation with 2 variables forgot the first 1 had a in it

Answer 20

but anyways this system should be solvable since we do have 3 equations with 3 unknowns
we just have to play around with it

Answer 21

alright! but wouldn't c=-b
?

Answer 22

write the equation should be 0=8b+8c
8b=-8c
b=-c

Answer 23

so that equation should say 0=-2b+2(-b)+1+a

Answer 24

0=-2b+1+a

Answer 25

2b-1=a

Answer 26

so you 2b-1=a and c=-b to write another equation in terms of just b

Answer 27

that will allow us to solve for b

Answer 28

you should get two possible values for b

Answer 29

have you gotten those yet

Answer 30

uhm im bit confused where to solve for b

Answer 31

use the equations given to you
for example you have 0=9b+6c-ba

Answer 32

replace a with (2b-1)
replace c with -b

Answer 33

\[\text{ solve } 0=9b+6(-b)-b(2b-1) \\ 0=9b-6b-2b^2+b \\ 0=4b-2b^2 \]

Answer 34

you will get two difference solutions for (a,b,c)
but we will see later only one solution will work of the two

Answer 35

i got b=0 and b=-1/2

Answer 36

b=-1/2?

Answer 37

let me see

Answer 38

i think there was a mistake when subbing in b=-c into 2b-2c-a-1=0, a should be equal to a=1+4b

Answer 39

\[0=2b(2-b) \\ b=0 \text{ or } 2-b=0 \]

Answer 40

In the step: 0=-2b+2(-b)+1+a... 0=-4b+1+a

Answer 41

not, 2b-1=a

Answer 42

so a=4b-1

Answer 43

ok you can correct it but the process is still the same
we need to replace the a and c in terms of b

Answer 44

alright!

Answer 45

we can both do it
let me know what you get and we can compare our answers

Answer 46

would I substitute my b-values into b=-c to get c, and the 2b-2c-a-1=0 to get a?

Answer 47

we have b=-c
and you said a=4b-1

Answer 48

so we just plug our b's into there to get our c and a

Answer 49

did you get (a,b,c)=(3,1,-1)?

Answer 50

my solution is backwards...
my max is at (-1,-2)
and my min is at (3,6)
:(

Answer 51

I will go back through it one more time to see if I could have done anything differently

Answer 52

I get, a =-1, b=0, c-0 or b=3/4, a=2, and -3/4=c

Answer 53

\[y=\frac{x^2+a}{bx+c} \\ y'=\frac{2x(bx+c)-b(x^2+a)}{(bx+c)^2} \\ y'=\frac{2bx^2+2cx-bx^2-ba}{(bx+c)^2} =\frac{bx^2+2cx-ba}{(bx+c)^2} \\ \text{we will get three equations} \\ \text{ one given by } f(-1)=-2 \text{ : } -2=\frac{1+a}{-b+c} \\ -2=\frac{1+a}{-b+c} \text{ rewrite this as : } 2b-2c=1+a \\ 2b-2c-1=a \\ \text{ now we also have } f'(3)=0 \\ \frac{9b+6c-ba}{(3b+c)^2}=0 \\ \text{ rewriting } 9b+6c-ba=0 \\ f'(-1)=0 \text{ gives } \frac{1b-2c-ba}{(-b+c)^2}=0 \\ \text{ rewriting } b-2c-ba=0 \\ \text{ so the three equations are : } \\ 2b-2c-1=a \\ 9b+6c-ba=0 \\ 1b-2c-ba=0 \\
\text{ So bottom two equations when subtracting one from the other gives } \\ 8b+8c=0 \\ b=-c \\
\text{ using } b=-c
\text{ for first equation } 2b-2c-1=a
\\ \text{ gives } \\ 2b+2b-1=a \text{ or } 4b-1=a \\
\text{ so let's look at the equation } 9b+6c-ba=0 \text{ and rewrite in terms of b } \\ 9b+6(-b)-b(4b-1)=0 \\ 9b-6b-4b^2+b=0 \\ 3b-4b^2+b=0 \\ 4b-4b^2 =0 \\ b=1 \text{ or } b=0 \\
\text{ so the two possibilities are : }
(a,b,c)=(4(1)-1),1,-(1))=(4-1,1,-1) \\ =(3,1,-1) \\ \text{ or } (a,b,c)=(4(0)-1,0,-(0))=(-1,0,0)
\\ \text{ and } (-1,0,0) \text{ makes no sense }
\]

Answer 54

http://www.wolframalpha.com/input/?i=f%28x%29%3D%28x%5E2%2B3%29%2F%28x-1%29
but this gives us what we wanted backwards if you see what I mean

Answer 55

yeah i understand, why (3,1,-1) make sense cause the graph gives you the local max and min from the orginal question

Answer 56

and the other option doesnt..

Answer 57

well it give us the min and max backwards though :(

Answer 58

awww

Answer 59

your question says max at x=3 and min at x=-1
but this solution gives us max at x=-1 and min at x=3

Answer 60

@wio what do you think here?

Answer 61

We got (3,1,-1) but our max and min is all mixed up

Answer 62

Kinda wonder if there is a mistake in the question

Well there could be a mistake in my work. My math is sloppy today but you pretty much corrected me on any stupid mistakes.

Answer 63

Hmmm

Answer 64

I didn't do anything with the second derivative because we already had 3 equations

Answer 65

So your equations are: \[
f(-1)=2\\
f'(3)=0 \land f'(-1)=0\\
f''(3)<0\land f''(-1)>0
\]

Answer 66

with 3 unknowns

Answer 67

Yeah, that is guaranteed for linear equations, but not for every type of equation.

Answer 68

To check my solutions I used wolfram's 3 equation system solver.
There seems to be no other possible solution.

Answer 69

Don't worry about it... I think there was a problem with the question, cause this is only first year calc, so i don;t think it would be that difficult. I'll just ask my prof.