Urgente help checking calculus 2 exercise please

[tatibarrera231]

Hi, I've been trying to solve this calculus ll exercise and I have been having problems, I wanted to know if you can examine it and tell me what you think, Is it wrong? Is it right?, What am I doing wrong?, How you would do it?, anything helps. Hope you can help and thank you very much in advance

1. Is continuous and derivable for the function in , determine its absolute ends and use the "Comparison Property" to find upper and lower dimensions for the area of the region enclosed by the graph of f and the x-axis, in the given interval

How I resolved the problem, please tell me how I did:


And I also had a few extra questions if anyone can help me understand;
1. I wanted to know if it would be correct to say that as f'(x) ≠ ∀ x ∈ (1/2, 3/2) and also f'(x) < 0 ∀ x ∈ (1/2, 3/2), then the function is decreasing and consequently the maximum value is reached at the lower limit of the interval, just as the minimum is reached at the upper limit?
2. And if the decreasing function in that interval is the slope of the tangent line for each value within that interval would be negative, therefore the derivative of f(x) would be negative?
3. And if you have to place the function between its absolute ends and apply the property to find the dimensions? And how would it be done?


I will be attentive to any help you guys can give me and I thank you very much in advance for your time even if you do not help me I am very grateful for taking the time to read the message.

 

[Dr.Peterson]

Is continuous and derivable for the function

in

, determine its absolute ends and use the "Comparison Property" to find upper and lower dimensions for the area of the region enclosed by the graph of f and the x-axis, in the given interval

First, we can clarify the translation. In proper English, I believe this would be

The function f(x) = cos(x)/x is continuous and differentiable in [1/2, 3/2]. Determine its absolute extrema (maximum and minimum), and use the comparison property to find upper and lower bounds for the area of the region enclosed by the graph of f and the x-axis in the given interval.​

Your work looks correct, though I would have wanted to show how you determined that the derivative is negative over that interval.

1. I wanted to know if it would be correct to say that as f'(x) ≠ ∀ x ∈ (1/2, 3/2) and also f'(x) < 0 ∀ x ∈ (1/2, 3/2), then the function is decreasing and consequently the maximum value is reached at the lower limit of the interval, just as the minimum is reached at the upper limit?

I presume you meant to say f'(x) ≠ 0 ∀ x ∈ (1/2, 3/2). This is correct, and appears to be what you actually did. Why are you unsure?

2. And if the decreasing function in that interval is the slope of the tangent line for each value within that interval would be negative, therefore the derivative of f(x) would be negative?

I'm not sure what you are asking. Some words appear to be missing. But possibly you are asking about the converse of the fact you used.

3. And if you have to place the function between its absolute ends and apply the property to find the dimensions? And how would it be done?

Taking "ends" to mean "extrema" and "dimensions" to mean "bounds", as above, this appears to refer to the "comparison property" you were told to use, and to what you did. Again, what are you unsure of? How is that property stated? (You may want to show us both the original language and the translation.)

 

[tatibarrera231]

Hi there, Thank you very much for your response

First, we can clarify the translation. In proper English, I believe this would be

The function f(x) = cos(x)/x is continuous and differentiable in [1/2, 3/2]. Determine its absolute extrema (maximum and minimum), and use the comparison property to find upper and lower bounds for the area of the region enclosed by the graph of f and the x-axis in the given interval.

Thank you

Your work looks correct, though I would have wanted to show how you determined that the derivative is negative over that interval.

I made this to demonstrate that as f'(x) ≠ ∀ x ∈ (1/2, 3/2) and also f'(x) < 0 ∀ x ∈ (1/2, 3/2), then the function is decreasing and consequently the maximum value is reached at the lower limit of the interval, just as the minimum is reached at the upper limit, Would that be correct?

(You may want to show us both the original language and the translation.)

I'm really sorry if it's not that clear, it was in Spanish

what are you unsure of? How is that property stated?

I was just unsure cause I have difficulties with calculus , I'm glad it was correct, I do have another question, I don't remember how the property I used here was called, if someone can help me, I'm trying to support my procedure, and I'm really bad at if if someone could help me please.

I will be attentive to any help you guys can give me and I thank you very much in advance again

 

[Dr.Peterson]

I made this to demonstrate that as f'(x) ≠ ∀ x ∈ (1/2, 3/2) and also f'(x) < 0 ∀ x ∈ (1/2, 3/2), then the function is decreasing and consequently the maximum value is reached at the lower limit of the interval, just as the minimum is reached at the upper limit, Would that be correct?

I did the same thing to check out your result; but you can't officially use a graph as proof.

I was just unsure cause I have difficulties with calculus , I'm glad it was correct, I do have another question, I don't remember how the property I used here

was called, if someone can help me, I'm trying to support my procedure, and I'm really bad at if if someone could help me please.

In order to be sure I'm answering your question, I have to repeat mine. Can you show me the statement of what you called the "comparison property"?

But also, the property you used within the clip you show is that the two cosines shown are positive. Is your question here about that, or about what you concluded from these facts?

 

[tatibarrera231]

I did the same thing to check out your result; but you can't officially use a graph as proof.

Okay, I understand, but then how can I demonstrate it? Do you know a way?

In order to be sure I'm answering your question, I have to repeat mine. Can you show me the statement of what you called the "comparison property"?

Of course, I'll leave the image here, sorry I didn't leave it earlier

Para la función en , determine sus extremos absolutos y use la "propiedad de comparcion" para encontrar cotas superior e inferior para el area de la region encerrada por el gráfico de f y el eje x, en el intervalo dado.

I left the image and the text written in case you want to translate a word, so you don't have to write it all.

But also, the property you used within the clip you show is that the two cosines shown are positive. Is your question here about that, or about what you concluded from these facts?

It was about that, I'm now trying to support my procedure by writing the things I did to get to my result.

And thank you for all the help, I'm really grateful

 

[Dr.Peterson]

1. I wanted to know if it would be correct to say that as f'(x) ≠ 0 ∀ x ∈ (1/2, 3/2) and also f'(x) < 0 ∀ x ∈ (1/2, 3/2), then the function is decreasing and consequently the maximum value is reached at the lower limit of the interval, just as the minimum is reached at the upper limit?

To show that f'(x) < 0 in that interval, you determine what f'(x) is, as you did, (-x sin(x) - cos(x))/x^2, and show algebraically that -x sin(x) - cos(x) < 0. Over the given interval, which is a subset of (0, pi/2), x and sin(x) and cos(x) are all positive, so ...

Of course, I'll leave the image here, sorry I didn't leave it earlier

Para la función

en

, determine sus extremos absolutos y use la "propiedad de comparcion" para encontrar cotas superior e inferior para el area de la region encerrada por el gráfico de f y el eje x, en el intervalo dado.

I left the image and the text written in case you want to translate a word, so you don't have to write it all.

It's nice to see the problem exactly as given, but I asked, not for the problem, but for the "comparison property" it refers to. Can you show us that in the same way?

Incidentally, Google translates the problem as

For the function ... in ..., determine its absolute extremes and use the "comparison property" to find upper and lower bounds for the area of the region enclosed by the graph of f and the x-axis, in the given interval.​

That's better than I expected! But I see the problem itself didn't say "continuous and differentiable"; I was curious how that fit in. I suppose you added it?