My Math Forum Find the coordinates of the point(s) at which the curve has the specified gradient

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 September 6th, 2017, 08:59 AM #1 Newbie   Joined: Sep 2017 From: Latvia/Denmark Posts: 21 Thanks: 0 Find the coordinates of the point(s) at which the curve has the specified gradient Hello. I come here because I am in need of help. I am desperate. I just started learning at a university, and I do not understand a lot of things here. I will just add a picture of what I do not understand. If someone would help me, it'd mean the world to me... I do not understand how to do the D, and I am stuck in C also... so, firstly I need the D, but if you would be so kind and explain the C also, I'd be really grateful. P.S. What would you suggest for a person who hasn't studied derivatives and slopes, but desperately needs to know this? Last edited by skipjack; September 6th, 2017 at 11:56 AM.
 September 6th, 2017, 12:24 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,149 Thanks: 1418 If the given equation is effectively $y = x^n$, where $n$ is a constant, the gradient is given by the derivative of $y$, which is $nx^{n-1}\!$. To learn about this, read any textbook that introduces calculus for beginners or use an online course. If you can't cope with the former, try the latter (but it may take longer if it includes watching lengthy videos). (d) Assuming that $x > 0$, you effectively have $y = x^{-3/2}\!$, so you need to solve $-{\small\dfrac32}x^{-5/2} = -\small\dfrac{3}{64}$, which simplifies to $x^{5/2} = 32$. That implies that $x^{1/2} = 2$, so $x = 4$. If you do (c) similarly, you should find you need to solve $-2x^{-3} = 16$, which simplifies to $x^3 = -1/8$. Can you solve that? Thanks from vizzy22 Last edited by skipjack; September 7th, 2017 at 02:39 AM.
September 7th, 2017, 01:33 AM   #3
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 Originally Posted by skipjack If the given equation is effectively $y = x^n$, where $n$ is a constant, the gradient is given by the derivative of $y$, which is $nx^{n-1}\!$. To learn about this, read any textbook that introduces calculus for beginners or use an online course. If you can't cope with the former, try the latter (but it may take longer if it includes watching lengthy videos). (d) Assuming that $x > 0$, you effectively have $y = x^{-3/2}\!$, so you need to solve $-{\small\dfrac32}x^{-5/2} = -\small\dfrac{3}{64}$, which simplifies to $x^{5/2} = 32$. That implies that $x^{1/2} = 2$, so $x = 4$. If you try to do (c) similarly, you should find you need to solve $-2x^{-3} = 16$, which simplifies to $x^3 = -1/8$. Can you solve that?
The thing is, how did you get $y = x^{-3/2}\!$ in D?
And in C, how do you get $x^3 = -1/8$ out of $-2x^{-3} = 16$, because for me it looks like $x^{-3} = -8$.

Last edited by skipjack; September 7th, 2017 at 02:56 AM.

 September 7th, 2017, 01:39 AM #4 Newbie   Joined: Sep 2017 From: Latvia/Denmark Posts: 21 Thanks: 0 Another thing - I was watching a video, and I had a question - why is the answer 1, not 1+1? Because I was wondering, if x' = 1, then where did that +1 go? Last edited by skipjack; September 7th, 2017 at 01:57 AM.
 September 7th, 2017, 02:42 AM #5 Global Moderator   Joined: Dec 2006 Posts: 18,149 Thanks: 1418 The function in (d) has denominator $x√x$, which is $x*x^{1/2}$, i.e. $x^{3/2}$. Note that $1/x^n = x^{-n}$. In (c), you get $x^{-3} = -8$, which is equivalent to $x^3 = -1/8$. The derivative of 1 (or any constant) is zero, so f(x) and f(x) + c (where c is a constant) have the same derivative, provided that f(x) is differentiable. This article gives a list of differentiation rules. Thanks from vizzy22
September 7th, 2017, 08:35 AM   #6
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 Originally Posted by skipjack The function in (d) has denominator $x√x$, which is $x*x^{1/2}$, i.e. $x^{3/2}$. Note that $1/x^n = x^{-n}$. In (c), you get $x^{-3} = -8$, which is equivalent to $x^3 = -1/8$. The derivative of 1 (or any constant) is zero, so f(x) and f(x) + c (where c is a constant) have the same derivative, provided that f(x) is differentiable. This article gives a list of differentiation rules.
Thank you, but on a different note. Do you think it is possible for me to catch up by studying on my own? In Latvia we weren't taught this in HS, yet in a university in Denmark, they are already expecting us to know this.

Last edited by skipjack; September 7th, 2017 at 02:16 PM.

 September 7th, 2017, 02:25 PM #7 Global Moderator   Joined: Dec 2006 Posts: 18,149 Thanks: 1418 It's possible, but could take quite a long time. During that time, you might fall behind on your university course, as it will be more difficult due to your lack of knowledge in various areas. Is the university course a mathematics course?
September 8th, 2017, 02:05 AM   #8
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 Originally Posted by skipjack It's possible, but could take quite a long time. During that time, you might fall behind on your university course, as it will be more difficult due to your lack of knowledge in various areas. Is the university course a mathematics course?
European studies, international relations and business, so basically, I have take maths for the "'business" part.

 September 8th, 2017, 03:06 AM #9 Global Moderator   Joined: Dec 2006 Posts: 18,149 Thanks: 1418 That helps. Do you consider yourself to be someone who can readily learn from textbooks by yourself? Last edited by skipjack; September 8th, 2017 at 04:31 AM.
September 8th, 2017, 04:23 AM   #10
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 Originally Posted by skipjack That helps. Do you consider yourself to be someone who can readily learn from textbooks by yourself?
I am quite stressed about this, because I barely passed my math course in Latvia, and I don't consider myself to excel in this field. I am, however, a real hard worker, so I hope I can push myself to the limit of understanding.

Last edited by skipjack; September 8th, 2017 at 04:32 AM.

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