Question

Let f(x) = (-3x^2 +16x +c) / (x^3 - 5x^2)
Determine teh constant c such that lim x->5 f(x) exists.

Answer 1

you want to make a hole here so that the graph jumps over the spot; as opposed to creating a vertical asymptote i believe

Answer 2

T = -3x^2 +16x +c; (-3x-B)(x-5)

D = x^3 - 5x^2; x^2 (x-5)

Answer 3

does that make sense? we are trying to control the offending denominator with its likeness up top

Answer 4

um what does t and d stand for?

Answer 5

"t"op and "d"enominator .... just names

Answer 6

oh ok

Answer 7

what we need to do is get it so (x-5) is a factor of the top part ...

Answer 8

-3x^2 +16x +c = (-3x-B) (x-5)

-3x-B
x-5
------
-3x^2 -Bx
+15x +15
---------------
-3x^2+16x +15

B=-1

Answer 9

..... let me try that again

Answer 10

-3x-B
x-5
------
-3x^2 -Bx
+15x +5B
---------------
-3x^2+16x +5B

B=-1; c=-5 is what i think i meant to say .. lets try it out

Answer 11

ok

Answer 12

http://www.wolframalpha.com/input/?i=%28-3x^2%2B16x-5%29%2F%28x^3-5x^2%29%2C+x%3D5

Answer 13

im not sure i understnad what this means

Answer 14

it means that the limit from the left and the right are the same when x approaches 5.

Answer 15

if we were not able to control that offending part of the denominator; there would be no limit at x=5

Answer 16

but how do we show work for this??

Answer 17

factor the bottom to see the bad part;

create it as a factor for the top; and solve

Answer 18

wait is teh wrk what u showed b4

Answer 19

yes, i try to work it the stated problem as a means to solve it ... lol

Answer 20

\[\lim_{x->5}\frac{-3x+16x+c}{x^3-5x^2}\]
\[\lim_{x->5}\frac{-3x+16x+c}{x^2(x-5)}\]
\[\lim_{x->5}\frac{(Ax+B)(x-5)}{x^2(x-5)}\]
where (Ax+B)(x-5) = -3x+16x+c

Answer 21

\[(Ax+B)(x-5) = -3x+16x+c\]
\[Ax^2+(-5A+B)x-5B= -3x+16x+c\]
and equate coefficients
\[Ax^2=-3x^2\]\[(-5A+B)x=16x\]\[-5B=c\]
\[A=-3\]\[-5(-3)+B=16;\ B=1\]\[-5(1)=c\]

Answer 22

wow i wud have never thought of this solution by myself. This makes so much more sense. If i have to find a constat but tehre is no denominaotr what shud i do?

Answer 23

if there is no denominator ..... well, other than it just being a polynomial, id have to have an example of what you mean.

Answer 24

it is because of the denominator here that we get a bad value at x=5

Answer 25

polynomials as a rule are continuous and need no control

Answer 26

here is an example: If g(x) = x^3 - x^2 + x , show taht a there exist a c at g(c) =10

Answer 27

that is given a polynomial and has no bad parts; you just have to solve it for f(x) = 10.

Answer 28

err... g(x) in this case

Answer 29

it aint pretty in the end, but it is real

Answer 30

http://www.wolframalpha.com/input/?i=x^3+-+x^2+%2Bx+%3D+10

Answer 31

is that supposed to b teh answer??

Answer 32

it looks waaay 2 confusing XD

Answer 33

i dont think that is the answer. but it helps to see that an answer exists.

now as i see it, they are not asking you to locate c, but to show that it exists

Answer 34

YES

Answer 35

so do we approach it differntly?

Answer 36

we can utilize the definition of a derivative:
\[\lim_{x->c}\frac{g(c)-g(x)}{c-x}\]

Answer 37

or \[\lim_{h->0}\frac{g(x+h)-g(x)}{h}\]
\[\lim_{h->0}\frac{(x+h)^3 - (x+h)^2 + (x+h)-(x^3 - x^2 + x)}{h}\]
\[\lim_{h->0}\frac{(x+h)^3 - (x^2+2xh+h^2) + x+h-x^3 + x^2 - x}{h}\]
\[\lim_{h->0}\frac{(x+h)^3 - x^2-2xh-h^2 +h-x^3 + x^2 }{h}\]
\[\lim_{h->0}\frac{(x+h)^3 -2xh-h^2 +h-x^3 }{h}\]
\[\lim_{h->0}\frac{(x^3+3x^2h+3xh^2+h^3) -2xh-h^2 +h-x^3 }{h}\]
\[\lim_{h->0}\frac{3x^2h+3xh^2+h^3 -2xh-h^2 +h }{h}\]
\[\lim_{h->0}\frac{h(3x^2+3xh+h^2 -2x-h +1) }{h}\]
\[\lim_{h->0}\ 3x^2+3xh+h^2 -2x-h +1\]
and when h=0 we get
\[3x^2 -2x +1\]
but this might be a different route

Answer 38

and if none of this looks familiar; then it prolly is another route that they are asking for

Answer 39

how does the 2nd way taht u showed prove taht g(c) = 10

Answer 40

this looks familiar..iots differnce qoutient

Answer 41

but im not sure i understand how it proves taht a c exists at g(c) =10

Answer 42

me either :) but then i havent got a clue as to what you are allowed to use in your proving

Answer 43

all polynomials are continuous, that is a rule somewhere

Answer 44

but are we trying to prove it by limits and deltas and epsilons or by some other means

Answer 45

oh either way it doesnt matter...ive seen both derivitives, dffnce quotient and delta n epsilons b4

Answer 46

http://mathrefresher.blogspot.com/2009/02/polynomials-are-continuous.html

Answer 47

this is "proof" but by what standards do we go by?

Answer 48

we are trying to prove it by limits moslt likely

Answer 49

since taht is what we have been doin in class

Answer 50

the epsilon-delta version starts out like this i think

g(c)-10 = e

0< |x-c| < d

Answer 51

<e not =e

Answer 52

how do we prove it by limits..if it is possible?

Answer 53

the limit of a sum is the sum of the limits

Answer 54

that might help to break it apart

Answer 55

lim(x^3-x^2+x) = lim(x^3) - lim(x^2)+lim(x)

Answer 56

but i got no idea how to prove it beyond htat

Answer 57

oh ok but do u know how 2 do it w/ epsilon n delta since taht is how u were doing it at first

Answer 58

its been to long to recall how to do it correctly; once you learn it, it tends to go by the wayside. id have to read up on it

Answer 59

oh ok thanks for ur help=)

Answer 60

yep, and good luck with it :)

Answer 61

thanks but can u help me w/ a dffncce quotient. it confuses me b/c terhe are negative exponets

Answer 62

post it up on the left and see who bites, ill get to it if i can

Answer 63

ok