Mathematics Question Thread

[gilliesb18]

Ahhhh can someone please help me understand domains and ranges??
And especially this one??
Find the range of each function over the given domain:
a) y= x^2 (x squared) for 0 < x < 3   (also i dont know how to make greater than or equal to signs, but in this question they are all greater than or equal to, if that makes sense)
Sorry about the muddly question too.....
Thanks!!!
B

[RuiAce]

Ahhhh can someone please help me understand domains and ranges??
And especially this one??
Find the range of each function over the given domain:
a) y= x^2 (x squared) for 0 < x < 3   (also i dont know how to make greater than or equal to signs, but in this question they are all greater than or equal to, if that makes sense)
Sorry about the muddly question too.....
Thanks!!!
B

Until you are experienced, you should never do such questions without a graph.

[gilliesb18]

Oh thanks soo much, it makes a lot of sense now!!!!
But I know what you mean about drawing a graph, especially as they get harder.
Thanks vv much.

[gilliesb18]

Hey I'm getting stuck again (not a suprise ) but can someone help me on this one now??
I have even drawn it up, i believe my parabola is correct, but where do i go from there??
So same question as before, Find the range of each function over the given domain, but this time, the question is;
y= -x^2 + 4 for -1 < x < 2

[RuiAce]

Hey I'm getting stuck again (not a suprise ) but can someone help me on this one now??
I have even drawn it up, i believe my parabola is correct, but where do i go from there??
So same question as before, Find the range of each function over the given domain, but this time, the question is;
y= -x^2 + 4 for -1 < x < 2

[JuliaPascale123]

Hey!! So for the first one, stationary points occur when \(x=-2\), \(f(x)=3\) (from the y-value), and \(f'(x)=0\) because it is a stationary point! Make those substitutions: into the function and the first derivative, to get a couple of equations:



So the function we now know is \(f(x)=x^3-12x-13\). Now it is just standard calculus. To find the turning points, put the 1st derivative equal to zero:



To find the inflexion, put the second derivative equal to zero:



To find the y-coordinates of any of these points, just pop the x-value back into the original function!

D is the hard bit - If you draw the curve you'll get something like this. What the question is asking is, where can you draw a horizontal line and cross that curve three times? That would mean you have three solutions to the equation \(f(x)=k\). For example, \(k=0\) works, because a horizontal line through \(y=0\) would cut the curve three times. The question is, what range of \(k\) values allow this to occur? At what point does it go too high or too low to cut the curve three times?

Hint: It is related directly to the coordinates of your turning points!

If that hint wasn't enough (pardon the edit Jamon!)

Specifically the y-coordinates

Hopefully this makes sense - Let me know if you need anything clarified

Hey,

I tried answering it but I have no idea what im doing, Would you mind telling me the answer please

[jamonwindeyer]

Hey,

I tried answering it but I have no idea what im doing, Would you mind telling me the answer please

No worries at all! So the line \(y=k\) will cut the curve three times as long as it sits above your minimum turning point, and below your maximum turning point. Outside this range is only cuts either once or twice the answer is



I'd recommend you try drawing some horizontal lines for \(y=-29\), \(y=3\), then values in between those and then not in between those! So you can spot the difference

[Fahim486]

Hi I'm having trouble with this question and sort of forgot how to do geometrical applications of calculus.
Thanks!

[RuiAce]

Hi I'm having trouble with this question and sort of forgot how to do geometrical applications of calculus.
Thanks!

[Fahim486]

Ahh I see that makes so much sense now. Thank you so much !!!!!!!

[bananna]

Hi, I'm having trouble with this q:

Use the formula Tn = ar^n−1 to find an expression for the 70th term of the GP with a=1andr=3

Also, I get really confused with the wording in these types of questions, like, do they want us to find Tn or n.

thanks!

[RuiAce]

Hi, I'm having trouble with this q:

Use the formula Tn = ar^n−1 to find an expression for the 70th term of the GP with a=1andr=3

Also, I get really confused with the wording in these types of questions, like, do they want us to find Tn or n.

thanks!

The 70th term is just \(T_{70}\)

Literally just substitute n=70, a=1 and r=30 straight into the formula to get \(3^{69}\) as your answer.

Look closely at the question. They clearly give you n.

[JuliaPascale123]

I'll copy paste the response with headings

Hey Julia! Welcome to the forums!

Question A
The first part is pretty straight forward, and I'm not sure how to construct a table on the forum. Basically, the quantity will decrease over time by 25% every 40 minutes. If t is minutes, and Q is in mg,







etc.

Question B

Now, we want to find an equation for Q. We know that what we're looking for here is some sort of exponential function. Once you've done these questions a number of times, you'll know that the form of the answer must be



Now, we know that the initial quantity is 300mg. So




We also know that the quantity has dropped to 225mg after 40 minutes. So,





So, the equation for the quantity of the medicine is



Question D (C Skipped)
Next, we want to find the time it takes for the quantity of medicine to half. I won't bother guessing and checking, let's just find the actual answer.





Remember that this is in minutes. So, dividing by 60, we get t=9.04 hours.

Question E
Part e starts to get tricky. We could use an arbitrary time t=a. However, it's exactly the same thing to start at an arbitrary quantity, A. That is because, at time t=a, there will be some quantity A of medicine in the system. Then, we would look for when A halves!



is our new formula. t is still arbitrary, but so is A. Now, we need to show that, regardless of A, the halving time of this equation is a constant (ie. the time we proved in part d).

Our halving time will occur when the quantity is 0.5A (half of the initial, arbitrary amount). So,




But this is exactly the same equation as above, with the same solution! So, the halving time is a constant for any arbitrary starting point.

Question F
The next part is just a standard sketch of an exponential function. If you're not sure what it will look like, plot some points, or use an online graphing tool.

Question G
For the last part, we solve for




Which equals 50.99 hours. We should definitely round up, for the safety of the patient, giving us 51 hours.

Graphing weebsites dont show the answer for g

[jakesilove]

Graphing weebsites dont show the answer for g

We can do a quick sanity check to see if G was correct;



Let's set A=1



Now, we sub in 3059.65 (our answer)



Which is what we would expect! Remember, we were looking for 0.02*A which, in this case is 0.02*1=0.02.

[itssona]

heeey

log base 2 of x is less than 3

solve

so i know log is defined for x>0 but idk how to do this, pls reply within an hour if u can before my exam aha, thank you