Question about function properties

revisisland24

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Hey guys,

I'm working on a review packet for my calc final and was wondering if my answers are correct. Here's the question:

Create a function g which satisfies the following criteria:

lim g(x) as x approaches negative infinity = infinity
lim g(x) as x approaches (-2) = 4
g is not continuous at -2
lim g(x) as x approaches infinity = 0
g(4) is undefined

my answer is (-12x/x-4) - that satisfies two of the requirements but I'm unsure how to satisfy the rest. Thanks!
 
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Hey guys,

I'm working on a review packet for my calc final and was wondering if my answers are correct. Here's the question:

Create a function g which satisfies the following criteria:

lim g(x) as x approaches infinity = infinity How can this be so if
lim g(x) as x approaches (-2) = 4
g is not continuous at -2
lim g(x) as x approaches infinity = 0 this is so. Contradiction: zero does not equal infinity.
g(4) is undefined

my answer is (-12x/x-4) - that satisfies two of the requirements but I'm unsure how to satisfy the rest. Thanks!
Please check the statement of the problem. There seems to be a contradiction.

Are you studying differential or integral calculus?

And which two criteria do you believe your proposed function satisfies?
 
Please check the statement of the problem. There seems to be a contradiction.

Are you studying differential or integral calculus?

And which two criteria do you believe your proposed function satisfies?

I'm sorry the top one should say "as x approaches negative infinity" i'm going to change it. and I believe I satisfied the limit as x approaches -2 and the g(4) is undefined
 
I'm sorry the top one should say "as x approaches negative infinity" i'm going to change it. and I believe I satisfied the limit as x approaches -2 and the g(4) is undefined
I agree about g(4). That is certainly undefined for your function.

\(\displaystyle \displaystyle \lim_{x \rightarrow - 2}g(x) = \lim_{x \rightarrow - 2}\left(\dfrac{-12x}{x - 4}\right) = \dfrac{\displaystyle \lim_{x \rightarrow - 2}(-12x)}{\displaystyle \lim_{x \rightarrow - 2}(x - 4)} = \dfrac{(- 12) * (- 2)}{(-2) - 4} = \dfrac{24}{- 6 } = - 4 \ne 4.\)

What is a common kind of function with a limit of a as x approaches b, but that is not continuous at a? What does that even mean?

What is a common kind of function with a limit of zero as x approaches plus or minus infinity?

What is a common kind of function with a limit of infinity as x approaches plus or minus infinity?

Hint: Think about functions that have a denominator of \(\displaystyle x^2 + 2x - 8.\)
 
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((16-xx)e^(-x-4))/2/(x+4)
this is not disc at -2, but gives some ideas
 
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I agree about g(4). That is certainly undefined for your function.

\(\displaystyle \displaystyle \lim_{x \rightarrow - 2}g(x) = \lim_{x \rightarrow - 2}\left(\dfrac{-12x}{x - 4}\right) = \dfrac{\displaystyle \lim_{x \rightarrow - 2}(-12x)}{\displaystyle \lim_{x \rightarrow - 2}(x - 4)} = \dfrac{(- 12) * (- 2)}{(-2) - 4} = \dfrac{24}{- 6 } = - 4 \ne 4.\)

What is a common kind of function with a limit of a as x approaches b, but that is not continuous at a? What does that even mean?

What is a common kind of function with a limit of zero as x approaches plus or minus infinity?

What is a common kind of function with a limit of infinity as x approaches plus or minus infinity?

Hint: Think about functions that have a denominator of \(\displaystyle x^2 +6x + 8.\)

I made a piecewise function where g(x) = 2x+2 when g<-2 and g(x) = 1/(x-4) when g>-2. Does that satisfy the criteria?
 
I made a piecewise function where g(x) = 2x+2 when g<-2 and g(x) = 1/(x-4) when g>-2. Does that satisfy the criteria?
A piecewise function certainly seems a sensible way to go.

Let's consider \(\displaystyle x < - 2 \implies g(x) = 2x + 2\ and\ x > - 2 \implies g(x) = \dfrac{1}{x - 4}.\)

Create a function g which satisfies the following criteria: lim g(x) as x approaches negative infinity = infinity

No, your proposed function does not meet this criterion: \(\displaystyle \displaystyle \lim_{x \rightarrow - \infty}(2x + 2) = - \infty \ne \infty.\)

lim g(x) as x approaches (-2) = 4

\(\displaystyle \displaystyle \lim_{x \rightarrow -2^-}(2x + 2) = 2(-2) + 2 = - 4 + 2 = -2 \ne 4.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow -2^+}\left(\dfrac{1}{x - 4}\right) = \dfrac{1}{- 2 - 4} = - \dfrac{1}{6} \ne 4.\)

So \(\displaystyle \displaystyle \lim_{x \rightarrow - 2}g(x)\ is\ undefined.\)

g is not continuous at -2
This is true because neither the function nor its limit are defined at x = - 2.

lim g(x) as x approaches infinity = 0
Yes.

g(4) is undefined
Yes

So you meet three of the five criteria.
 
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The key piece to solving this problem using a piecewise function is to start with the definitions of (g(x) at x above and below - 2. The limits from above and below must equal 4. Now many functions do that so you must further restrict the definitions, but at least the most difficult criterion will have been met. Also, consider what are the requirements for a rational function to have a limit at infinity of zero.
 
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A piecewise function certainly seems a sensible way to go.

Let's consider \(\displaystyle x < - 2 \implies g(x) = 2x + 2\ and\ x > - 2 \implies g(x) = \dfrac{1}{x - 4}.\)


No, your proposed function does not meet this criterion: \(\displaystyle \displaystyle \lim_{x \rightarrow - \infty}(2x + 2) = - \infty \ne \infty.\)



\(\displaystyle \displaystyle \lim_{x \rightarrow -2^-}(2x + 2) = 2(-2) + 2 = - 4 + 2 = -2 \ne 4.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow -2^+}\left(\dfrac{1}{x - 4}\right) = \dfrac{1}{- 2 - 4} = - \dfrac{1}{6} \ne 4.\)

So \(\displaystyle \displaystyle \lim_{x \rightarrow - 2}g(x)\ is\ undefined.\)


This is true because neither the function nor its limit are defined at x = - 2.


Yes.


Yes

So you meet three of the five criteria.

Thanks so if i make the first function g(x) = -2x and the other g(x) = -24/(x-4) it should satisfy all of them
 
Thanks so if i make the first function g(x) = -2x and the other g(x) = -24/(x-4) it should satisfy all of them
Certainly one solution to the problem is:

\(\displaystyle g(x) = -2x\ if\ x < -2,\ and\ g(x) = \dfrac{-24}{x - 4}\ if\ x > - 2.\)

Proof

1 lim g(x) as x approaches negative infinity = infinity

\(\displaystyle \displaystyle \lim_{x \rightarrow - \infty}g(x) = \lim_{x \rightarrow - \infty}(-2x) = \infty.\)

2 lim g(x) as x approaches (-2) = 4

\(\displaystyle \displaystyle \lim_{x \rightarrow - 2^-}g(x) = \lim_{x \rightarrow - 2^-}(-2x) = (-2)(-2) = 4.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow - 2^+}g(x) = \lim_{x \rightarrow - 2^+}\dfrac{-24}{-2 - 4} = \dfrac{-24}{-6} = 4.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow - 2}g(x) = 4.\)

3 g is not continuous at -2

You have not defined g(x) at x = - 2 so it does not exist there and cannot be continuous.

4 lim g(x) as x approaches infinity = 0

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}g(x) = \lim_{x \rightarrow \infty}\dfrac{-24}{x - 4} = \dfrac{\displaystyle \lim_{x \rightarrow \infty}(-24)}{\displaystyle \lim_{x \rightarrow \infty}(x - 4)} = \dfrac{-24}{\infty} = 0.\)

5 g(4) is undefined

\(\displaystyle \dfrac{1}{x - 4}\ is\ undefined\ at\ x = 4.\)

Well done.
 
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