Solving Integral & Sign Charts

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three3985

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Hello all,

I would like to check if my answers to the below questions are correct.

1618052009960.png

1618052022157.png

1618052055981.png

1618052115898.png

For 2a), I'm not exactly sure how to draw the sign chart because i'm unable to derive a quadratic equation such as (X+1)(X-3) which will allow me to begin drawing the sign chart.

Thanks for taking the time to read my questions!
 
three3985 said:
Hello all,

I would like to check if my answers to the below questions are correct.

View attachment 26358

View attachment 26359

View attachment 26360

View attachment 26361

For 2a), I'm not exactly sure how to draw the sign chart because i'm unable to derive a quadratic equation such as (X+1)(X-3) which will allow me to begin drawing the sign chart.

Thanks for taking the time to read my questions!
Click to expand...
post ONE problem per thread.

Your work for the first problem is correct.

For the second problem, I would sketch the function s first

Then find the points of intersection - which will define the area of interest and continue.....
 
Subhotosh Khan said:
post ONE problem per thread.

Your work for the first problem is correct.

For the second problem, I would sketch the function s first

Then find the points of intersection - which will define the area of interest and continue.....
Click to expand...

Apologies for posting two problems.

1618055005434.png

Would this be correct? I drew the sign chart based on the fact that the only intersection occur at (0,3) and the area of interest are both concave upwards.
 
In (a) you the correct answer but have an incorrect notation in the solution
You have \(\displaystyle ln|v|_2^8\) which is correct and immediately below it \(\displaystyle ln|x- 1|_2^8\) which is incorrect.
If you had actually calculated \(\displaystyle ln|x- 1|_2^8= ln|7|- ln|1|= ln(7)\) you would have the wrong answer,. But you didn't, you calculate \(\displaystyle ln|v|_2^8= ln(8)- ln(2)= ln(8/2)= ln(4)= ln(2^2)= 2 ln(2)\) which is correct.
 
HallsofIvy said:
In (a) you the correct answer but have an incorrect notation in the solution
You have \(\displaystyle ln|v|_2^8\) which is correct and immediately below it \(\displaystyle ln|x- 1|_2^8\) which is incorrect.
If you had actually calculated \(\displaystyle ln|x- 1|_2^8= ln|7|- ln|1|= ln(7)\) you would have the wrong answer,. But you didn't, you calculate \(\displaystyle ln|v|_2^8= ln(8)- ln(2)= ln(8/2)= ln(4)= ln(2^2)= 2 ln(2)\) which is correct.
Click to expand...
Hey, thanks for spotting that out! I'll amend it. Cheers!
 
Your graph shows that [MATH][/MATH][MATH] y=x^2-2x+3[/MATH] doesn't cross the [MATH]x[/MATH]-axis. (If it crossed the [MATH]x[/MATH]-axis, that would be a point where the two original graphs meet).
It is always above the [MATH]x[/MATH]-axis, so the graph of [MATH]y=x^2+1[/MATH] is always above the graph of [MATH]y=2x-2[/MATH]. Therefore, there's no cross-over, and in particular none in the interval [-1,2]
 
lex said:
Your graph shows that [MATH][/MATH][MATH] y=x^2-2x+3[/MATH] doesn't cross the [MATH]x[/MATH]-axis. (If it crossed the [MATH]x[/MATH]-axis, that would be a point where the two original graphs meet).
It is always above the [MATH]x[/MATH]-axis, so the graph of [MATH]y=x^2+1[/MATH] is always above the graph of [MATH]y=2x-2[/MATH]. Therefore, there's no cross-over, and in particular none in the interval [-1,2]
Click to expand...
Thanks for your response. Since the graph does not cross the x axis, then am I right to say that partition number is 0 for this case?
 
I'm not sure what you mean by 'partition number'. If you mean you can integrate without partitioning the interval [-1, 2], you are correct.
 
three3985 said:
Apologies for posting two problems.

View attachment 26362

Would this be correct? I drew the sign chart based on the fact that the only intersection occur at (0,3) and the area of interest are both concave upwards.
Click to expand...
You have two functions but only graphed one. Why is that?

is your chart for f(x), f'(x), f"(x) or something different? Seriously, how is one to know if your chart is correct if you fail to label it?
 
Jomo said:
You have two functions but only graphed one. Why is that?

is your chart for f(x), f'(x), f"(x) or something different? Seriously, how is one to know if your chart is correct if you fail to label it?
Click to expand...
Thanks for your response.

I don't see what's wrong with my graph. I keyed my solution into symbolab and it yielded the same result.

1618074926183.png


I was trying to plot the sign chart of f(x) - g(x) = x^2-2+3

Apologies for not labelling. I'm still trying to figure out this particular chapter.
 
Last edited:
lex said:
I'm not sure what you mean by 'partition number'. If you mean you can integrate without partitioning the interval [-1, 2], you are correct.
Click to expand...

If I present my graph as this highlighting the area of interest

1618075698119.png

and my present my sign chart of f(x) - g(x) = x^2-2+3 like this, I would be correct?

1618075749312.png
 
Oh, your graph is for f(x)-g(x)
Subhotosh Khan said:
post ONE problem per thread.

Your work for the first problem is correct.

For the second problem, I would sketch the function s first

Then find the points of intersection - which will define the area of interest and continue.....
Click to expand...
The area of interest was given. Graphing would help for the obvious reasons.
 
three3985 said:
If I present my graph as this highlighting the area of interest
...
and my present my sign chart of f(x) - g(x) = x^2-2+3 like this, I would be correct?
...
Click to expand...
1618081467461.png

It is clear from the graph that f(x) - g(x) is always > 0, so there's no need for the 'sign chart'; although yes it is + everywhere.
A 'sign chart' is only of practical use if the graph crosses the x-axis somewhere:
E.g. for the graph of a different function:
1618081316753.png

1618081345452.png
 
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