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RuiAce

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« Reply #1200 on: January 12, 2017, 11:06:02 pm »
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Thank you for your help jamon. Can i please have help with these two questions?
Look at Q9 again. There's an x in the numerator as well. You can't just chain rule it.

You have to either directly use the quotient rule, or use the product rule on your line 1.

Same goes for Q10. Your points of intersection are correct though
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anotherworld2b

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« Reply #1201 on: January 12, 2017, 11:40:35 pm »
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I see now
thank you RuiAce
Look at Q9 again. There's an x in the numerator as well. You can't just chain rule it.

You have to either directly use the quotient rule, or use the product rule on your line 1.

Same goes for Q10. Your points of intersection are correct though

anotherworld2b

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« Reply #1202 on: January 13, 2017, 12:44:34 am »
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Hi I'm back i just wanted to get some help with this question. I am not sure how to draw it... All i know that the gradient is either negative or positive at different points.

RuiAce

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« Reply #1203 on: January 13, 2017, 12:52:34 am »
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Hi I'm back i just wanted to get some help with this question. I am not sure how to draw it... All i know that the gradient is either negative or positive at different points.
You started it off well. I can see the faint +'s and -'s floating around.

Now the idea is, wherever you drew the +'s, the graph is above the x-axis for f'(x). Remember that f'(x) IS supposed to measure the gradient.

Similarly, where you drew -'s, they are below the x-axis for f'(x). Where you marked a 0, it becomes an x-intercept.
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bsdfjnlkasn

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« Reply #1204 on: January 13, 2017, 09:50:18 am »
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You started it off well. I can see the faint +'s and -'s floating around.

Now the idea is, wherever you drew the +'s, the graph is above the x-axis for f'(x). Remember that f'(x) IS supposed to measure the gradient.

Similarly, where you drew -'s, they are below the x-axis for f'(x). Where you marked a 0, it becomes an x-intercept.

Sorry, I was just looking on at this question and was wondering what happens in b) at the horizontal point of inflexion?
Because the gradient is 0 there would we bring f'(x) to approach 0 as though y = 0 were an asymptote (after the 3rd root)?
Or is this completely wrong?

Anyway, I hope my question makes some sort of sense

jamonwindeyer

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« Reply #1205 on: January 13, 2017, 09:55:59 am »
+1

Sorry, I was just looking on at this question and was wondering what happens in b) at the horizontal point of inflexion?
Because the gradient is 0 there would we bring f'(x) to approach 0 as though y = 0 were an asymptote (after the 3rd root)?
Or is this completely wrong?

Anyway, I hope my question makes some sort of sense

Almost right, but it would actually touch the y-axis! It actually turns back around on itself - You'll draw a turning point that just touches the x-axis,  if that makes sense? this is because the gradient goes from positive, to 0, back to positive (or replace positive with negative, depending on the function). This corresponds to the curve coming towards the axis, just touching it, then turning back around

RuiAce

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« Reply #1206 on: January 13, 2017, 10:01:08 am »
+1

Sorry, I was just looking on at this question and was wondering what happens in b) at the horizontal point of inflexion?
Because the gradient is 0 there would we bring f'(x) to approach 0 as though y = 0 were an asymptote (after the 3rd root)?
Or is this completely wrong?

Anyway, I hope my question makes some sort of sense
$\text{Aside of Jamon's explanation, I'll illustrate with an example}$
$y=x^3\\ \frac{dy}{dx}=3x^2$
$\text{The original function has a horizontal point of inflexion at }(0,0)\\ \text{So the derivative will ALSO have an }x\text{-intercept at }x=0...$
$...\text{but the }x\text{-intercept is ALSO a stationary point}\\ \text{From your studies in polynomials}\\ \text{You should be able to note that }\frac{dy}{dx}=3x^2\\ \text{has a 'double root' at }x=0$
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anotherworld2b

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« Reply #1207 on: January 13, 2017, 02:04:51 pm »
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I was working on this question here but i am not sure what i did wrong. The only part that i got wrong was determine whether each staionary point was a maximum, minimum or horizontal inflection?
« Last Edit: January 13, 2017, 02:10:08 pm by anotherworld2b »

jamonwindeyer

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« Reply #1208 on: January 13, 2017, 02:42:30 pm »
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I was working on this question here but i am not sure what i did wrong. The only part that i got wrong was determine whether each staionary point was a maximum, minimum or horizontal inflection?

Almost! Near the bottom, you've differentiated the wrong expression to get your second derivative. You need to differentiate $y=1-\frac{4}{(x+3)^2}$, not the rearranged version!

Besides that your method is spot on. Nice job!

anotherworld2b

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« Reply #1209 on: January 13, 2017, 05:06:47 pm »
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I see now
I am also having the same trouble with this question which is similar to q7
Almost! Near the bottom, you've differentiated the wrong expression to get your second derivative. You need to differentiate $y=1-\frac{4}{(x+3)^2}$, not the rearranged version!

Besides that your method is spot on. Nice job!

jakesilove

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« Reply #1210 on: January 13, 2017, 06:06:24 pm »
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I see now
I am also having the same trouble with this question which is similar to q7

When you've subbed points in to check whether the value is a POI etc. you've used the second derivative, instead of the first.The second derivative tells you about the concavity, not the gradient. So, you've shown that the concavity changes, which suggests a POI. Instead, you should sub the values (1/4 and 3/4) into the first derivative! This will tell you about changing gradients.
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anotherworld2b

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« Reply #1211 on: January 13, 2017, 07:53:53 pm »
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that makes more sense now

When you've subbed points in to check whether the value is a POI etc. you've used the second derivative, instead of the first.The second derivative tells you about the concavity, not the gradient. So, you've shown that the concavity changes, which suggests a POI. Instead, you should sub the values (1/4 and 3/4) into the first derivative! This will tell you about changing gradients.

hanaacdr

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« Reply #1212 on: January 14, 2017, 06:48:42 am »
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Hi can i please get some help on this Simple Harmonic question plz
I have attached it to this post

thank you
Much appreciated
« Last Edit: January 14, 2017, 06:52:30 am by hanaacdr »

RuiAce

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« Reply #1213 on: January 14, 2017, 09:52:54 am »
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Hi can i please get some help on this Simple Harmonic question plz
I have attached it to this post

thank you
Much appreciated
Before doing these questions, have you been taught how to do HSC (2U) trigonometric functions AS WELL as the 2U motion questions?
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