1. ## find x-coordinate

Hi:
I am asked to find the x-coordinate of B from y=12x⋀½ - 2x⋀3/2
I am sure this requires finding the turning points by differentiation dy/dx on the equation which gives 6x⋀-1/2 - 3x⋀1/2.
But how do I factorize this to find the x-coordinate? Or is my thinking wrong?
Thanks

⋀ - means raised to the power of

2. Originally Posted by lemonz
Hi:
I am asked to find the x-coordinate of B from y=12x⋀½ - 2x⋀3/2
I am sure this requires finding the turning points by differentiation dy/dx on the equation which gives 6x⋀-1/2 - 3x⋀1/2.
But how do I factorize this to find the x-coordinate? Or is my thinking wrong?
Thanks

⋀ - means raised to the power of
What's B meant to be? Turning point?

$\frac{6}{\sqrt{x}} - 3 \sqrt{x} = 0 \Rightarrow \frac{2}{\sqrt{x}} = \sqrt{x}$

Multiply both sides by $\sqrt{x}$:

$\Rightarrow 2 = x$.

3. ## re: indices

Hi: Thanks:
yes, B is a max. turning point.
I am still having trouble getting x=2 from the given equation - out of standard index form. Your answer was correct. X does = 2.
Could you show me the interim steps, please?
regards

4. Originally Posted by lemonz
Hi: Thanks:
yes, B is a max. turning point.
I am still having trouble getting x=2 from the given equation - out of standard index form. Your answer was correct. X does = 2.
Could you show me the interim steps, please?
regards
Mr. Fantastic did show the intermediate steps...

$y=12x^{1/2} - 2x^{3/2}$

$\frac{dy}{dx} = 6x^{-1/2} - 3x^{1/2} = 0$

And here is his work, following immediately from where you left off:
$\frac{6}{\sqrt{x}} - 3 \sqrt{x} = 0$

$\frac{2}{\sqrt{x}} = \sqrt{x}$

$2=x$

Which part did you miss?

Hi:
I don't understand how Mr.Fantastic got from 6 to 2 with minus of 3
thanks

6. $\frac{6}{\sqrt{x}} - 3 \sqrt{x} = 0$

$3\left(\frac{2}{\sqrt{x}} - \sqrt{x}\right) = 0$

$\frac{2}{\sqrt{x}} - \sqrt{x} = 0$

$\frac{2}{\sqrt{x}} = \sqrt{x}$

$2=x$

7. Originally Posted by skeeter
$\frac{6}{\sqrt{x}} - 3 \sqrt{x} = 0$

$3\left(\frac{2}{\sqrt{x}} - \sqrt{x}\right) = 0$

$\frac{2}{\sqrt{x}} - \sqrt{x} = 0$

$\frac{2}{\sqrt{x}} = \sqrt{x}$

$2=x$
My mistake for assuming a calculus student will know how to divide out a common factor (of 3 in this case).

8. Sad to say, but my experience has been such that I will never assume anything about a calculus student's algebra (or geometry) skills.

9. Honestly, after moving on to more advanced math, I tend to lose some of the earlier skills that I had.

For example, I can no longer multiply two-digit numbers in my head. I can't even count properly anymore. All I do these days is count from 1 to n, or worse yet, enumerate from 1 to aleph-0

LOL

Math is never trivial. Hindsight is always 20-20, but learning math for the first time can be quite difficult. I can't tell you how many times I asked my friends a question and got smacked by with "Dude, that's trivial! You can solve by inspection - you don't see it? Nooo?"

10. ## Calculus students

Sorry.
I'm not a calculus student. I'm just taking my school exams. I didn't realize you had to be a calculus student in order to post calculus based questions in this forum. My mistake.

11. Originally Posted by lemonz
Sorry.
I'm not a calculus student. I'm just taking my school exams. I didn't realize you had to be a calculus student in order to post calculus based questions in this forum. My mistake.
If you're learning about differentiation and its applications then you're studying calculus and so you're a calculus student. There are certain mathematical skills that a calculus student is assumed to have. One of those skills is algebraic skills such as dividing out a common factor in an equation. That is what was being discussed, not whether or not you were allowed to post a calculus question.

Since the calculus forum is obviously a forum for asking questions about calculus the question you asked was quite rightly posted in this forum.

If you are preparing for exams it might be wise to revise and consolidate the background algebraic skills the examiner will assume that you have.

In the meanwhile your question has been answered and it's probably best that the thread is closed.