Find the points on a graph at which the tangent line is parallel

  • #1

Homework Statement


Find the points on the graph y=x^3/2 - x^1/2 at which the tangent line is parallel to y-x=3.


Homework Equations





The Attempt at a Solution


First I found that the derivative of y=x^3/2 - x^1/2 is 1x.
I then rewrote the other line as y = 3+x and found the derivative to be 1.
Then I set 1x = 1, therefore x = 1
Now I plugged it back into the original function and got y = 1^3/2-1^1/2 = 1
So the point is (1,1).
But I have no idea if I am anywhere near correct. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
35,017
6,767

Homework Statement


Find the points on the graph y=x^3/2 - x^1/2 at which the tangent line is parallel to y-x=3.


Homework Equations





The Attempt at a Solution


First I found that the derivative of y=x^3/2 - x^1/2 is 1x.
No it isn't. How did you get 1x for the derivative?
I then rewrote the other line as y = 3+x and found the derivative to be 1.
Then I set 1x = 1, therefore x = 1
Now I plugged it back into the original function and got y = 1^3/2-1^1/2 = 1
So the point is (1,1).
But I have no idea if I am anywhere near correct. Any help would be greatly appreciated.

Also, calculus questions should be posted in "Calculus & Beyond" not in the Precalc Math section. I am moving this thread.
 
  • #3
My apologies for the mispost. Also I do not know what I was thinking with that derivative. It should be 3x^1/2/2-1/2x^1/2 correct?
 
  • #4
Therefore, setting 3x^1/2/2-1/2x^1/2 = 1, I get x = 1/9, 1.
 
  • #5
35,017
6,767
My apologies for the mispost. Also I do not know what I was thinking with that derivative. It should be 3x^1/2/2-1/2x^1/2 correct?

Almost, and you need to use parentheses around the exponents if you write using ordinary text.

The first term, (3/2)x^(1/2), in your derivative is correct, but the second one isn't.
 
  • #6
35,017
6,767
Here's the first part of what you wrote: 3x^1/2/2

Using the precedence of operations, this is the same as if written ((3x^1)/2)/2

Or, in LaTeX:
$$ \frac{\frac{3x^1}{2}}{2} = \frac{3x^1}{4}$$
 
  • #7
The second part of the derivative, (1/2)x^(-1/2) = (1)/2x^(1/2) is incorrect?
 
  • #8
Ah, perhaps it is (1)/2√x
 
  • #9
35,017
6,767
The second part of the derivative, (1/2)x^(-1/2) = (1)/2x^(1/2) is incorrect?

(1/2)x^(-1/2) is correct, but it's not equal to (1)/2x^(1/2).

Ah, perhaps it is (1)/2√x
This is also incorrect.

According to the order of operations, (1)/2x^(1/2) means
$$ \frac{1}{2}\cdot x^{1/2}$$

Similarly, (1)/2√x means
$$ \frac{1}{2}\cdot \sqrt{x}$$

If you write these using inline text, this will do it: 1/(2x^(1/2)) or 1/(2√x). Now it's clear what's in the denominator.
 
  • #10
Regardless of my incorrect use of inline text. Did I solve for x properly when I came to x= 1/9 and 1?
 
  • #11
35,017
6,767
If you substitute either of these values (x = 1/9, x = 1) into the derivative, do you get 1 as the result?
 
  • #12
x = 1 does, x = 1/9 seems to give me -1.
 
  • #13
Clearly I am doing something very wrong.
 
  • #15
(3x^(1/2))/2 - 1/(2x^(1/2)) = 1
(3x^(1/2))/2 · 2x^(1/2) - 1/(2x^(1/2)) · 2x^(1/2) = 1 · 2x^(1/2)
3x - x^0 = 2x^(1/2)
3x - 1 - 2x^(1/2) = 0
3u^2 - 2u - 1 = 0 where u = x^(1/2)
(3u+1)(u-1) = 0

Now 3u+1 = 0
3u = -1
(3u)/3 = -1/3
u = -1/3
x^(1/2) = -1/3
x = (-1/3)^2
x = 1/9
 
  • #16
35,017
6,767
(3x^(1/2))/2 - 1/(2x^(1/2)) = 1
(3x^(1/2))/2 · 2x^(1/2) - 1/(2x^(1/2)) · 2x^(1/2) = 1 · 2x^(1/2)
3x - x^0 = 2x^(1/2)
3x - 1 - 2x^(1/2) = 0
3u^2 - 2u - 1 = 0 where u = x^(1/2)
(3u+1)(u-1) = 0

Now 3u+1 = 0
3u = -1
(3u)/3 = -1/3
u = -1/3
So x1/2 = -1/3, which means there is no solution for x. There is no real number whose (principal) square root is negative.
x^(1/2) = -1/3
x = (-1/3)^2
x = 1/9
 
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  • #17
Ah. So the only solution is x = 1, which I plug back into the original equation y = x^(3/2) - x^(1/2), and gives me 0? Therefore, the point at which the tangent line is parallel to y-x=3 is (1,0)?
 
  • #19
Thank you so much.
 

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