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

In summary: I really appreciate it!In summary, the points on the graph y=x^3/2 - x^1/2 at which the tangent line is parallel to y-x=3 are (1,0).
  • #1
thatguythere
91
0

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.
 
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  • #2
thatguythere said:

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?
thatguythere said:
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
thatguythere said:
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
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
thatguythere said:
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).

thatguythere said:
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
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.
 
  • #14
thatguythere said:
x = 1 does, x = 1/9 seems to give me -1.
x = 1 is a solution. How did you get x = 1/9?
 
  • #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
thatguythere said:
(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.
thatguythere said:
x^(1/2) = -1/3
x = (-1/3)^2
x = 1/9
 
Last edited:
  • #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.
 

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

What does it mean for a tangent line to be parallel to a graph?

When a tangent line is parallel to a graph, it means that the slope of the tangent line is equal to the slope of the graph at that point. This indicates that the two lines have the same direction and will never intersect.

How do you find the points on a graph where the tangent line is parallel?

To find these points, you first need to take the derivative of the function that represents the graph. Then, you set the derivative equal to the slope of the tangent line you are trying to find. Finally, solve for the x-values that satisfy the equation to determine the points at which the tangent line is parallel.

Can there be more than one point on a graph where the tangent line is parallel?

Yes, there can be multiple points on a graph where the tangent line is parallel. This occurs when the graph has a constant slope or when the slope of the tangent line is equal to the slope of the graph at multiple points.

What is the significance of finding points on a graph where the tangent line is parallel?

Finding these points allows you to determine the critical points of a graph, which are points where the graph changes from increasing to decreasing or vice versa. These points are important in optimization problems and can also help in understanding the overall behavior of a graph.

Are there any practical applications of finding points on a graph where the tangent line is parallel?

Yes, there are many practical applications of this concept. For example, in physics, finding these points can help determine the maximum or minimum values of a physical quantity. In economics, they can be used to find the optimal production level for a company. Additionally, understanding the behavior of a graph can assist in predicting trends in data analysis and forecasting.

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