Tangent line questions!
- Thread starter farealol
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(a) I would write:
[MATH]f(x)=x^{-\frac{1}{n}}[/MATH]
Thus:
[MATH]f'(x)=-\frac{1}{n}x^{-\frac{1}{n}-1}=-\frac{1}{n}x^{-\frac{n+1}{n}}[/MATH]
And so the tangent line will be:
[MATH]y=f'(1)(x-1)+1=-\frac{1}{n}(x-1)+1=-\frac{1}{n}x+\frac{n+1}{n}[/MATH]
Here's a live graph:
www.desmos.com
[MATH]f(x)=x^{-\frac{1}{n}}[/MATH]
Thus:
[MATH]f'(x)=-\frac{1}{n}x^{-\frac{1}{n}-1}=-\frac{1}{n}x^{-\frac{n+1}{n}}[/MATH]
And so the tangent line will be:
[MATH]y=f'(1)(x-1)+1=-\frac{1}{n}(x-1)+1=-\frac{1}{n}x+\frac{n+1}{n}[/MATH]
Here's a live graph:
Desmos | Graphing Calculator
www.desmos.com
Ok I got it. Thank you so much.(a) I would write:
[MATH]f(x)=x^{-\frac{1}{n}}[/MATH]
Thus:
[MATH]f'(x)=-\frac{1}{n}x^{-\frac{1}{n}-1}=-\frac{1}{n}x^{-\frac{n+1}{n}}[/MATH]
And so the tangent line will be:
[MATH]y=f'(1)(x-1)+1=-\frac{1}{n}(x-1)+1=-\frac{1}{n}x+\frac{n+1}{n}[/MATH]
Here's a live graph:
Desmos | Graphing Calculator
Ok thank you!
I noticed one sign mistake.. but besides that is there anything else wrong?
Steven G
Elite Member
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- Dec 30, 2014
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When you solve for x there should be no x on the other side.
Else it would be easy to solve for x in any expression. For example: 2x-7 = 4x^2. Then 4x^2-2x + 7 =0. Now add x to goth sides. x = 4x^2 -2x+7 +x. No, it does not work that way.
Another example. Solve for x in 3+x =0. so 3+x+x = x. That is x=3+x+x (I added x to 0 to get x on one side of the equal sign).
Else it would be easy to solve for x in any expression. For example: 2x-7 = 4x^2. Then 4x^2-2x + 7 =0. Now add x to goth sides. x = 4x^2 -2x+7 +x. No, it does not work that way.
Another example. Solve for x in 3+x =0. so 3+x+x = x. That is x=3+x+x (I added x to 0 to get x on one side of the equal sign).
Thank you and yes I understand that.
I tried to put all the terms with x on one side and terms with n on the other side, but when I was trying to solve the equation, I found that there is always some term with both x and n, while other terms only have either x or n.
All right I will try again. Thank you!