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

[thatguythere]

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.

 

[Mark44]

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.

 

[thatguythere]

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?

 

[thatguythere]

Therefore, setting 3x^1/2/2-1/2x^1/2 = 1, I get x = 1/9, 1.

 

[Mark44]

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.

 

[Mark44]

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:
$$ \frac{\frac{3x^1}{2}}{2} = \frac{3x^1}{4}$$

 

[thatguythere]

The second part of the derivative, (1/2)x^(-1/2) = (1)/2x^(1/2) is incorrect?

 

[thatguythere]

Ah, perhaps it is (1)/2√x

 

[Mark44]

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.

 

[thatguythere]

Regardless of my incorrect use of inline text. Did I solve for x properly when I came to x= 1/9 and 1?

 

[Mark44]

If you substitute either of these values (x = 1/9, x = 1) into the derivative, do you get 1 as the result?

 

[thatguythere]

x = 1 does, x = 1/9 seems to give me -1.

 

[thatguythere]

Clearly I am doing something very wrong.

 

[Mark44]

x = 1 does, x = 1/9 seems to give me -1.

x = 1 is a solution. How did you get x = 1/9?

 

[thatguythere]

(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

 

[Mark44]

(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 x^1/2 = -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

 

[thatguythere]

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)?

 

[Mark44]

Looks good.

 

[thatguythere]

Thank you so much.