# slope of tangent

• Sep 19th 2008, 03:21 PM
silencecloak
slope of tangent
$y=x-x^5$

find the slope of the tangent line to the curve at point (1,0)

slope =

$
\frac{f(x)-f(a)}{x-a}
$

I'm a little confused on what 'a' equals, but going off my notes im gonna say its 1, being x value in the point

$
\frac{(x-x^5)-(1-1^5)}{x-1}
$

=

$
\frac{x-x^5}{x-1}
$

and thats where I get stuck
• Sep 19th 2008, 03:26 PM
Jhevon
Quote:

Originally Posted by silencecloak
$y=x-x^5$

find the slope of the tangent line to the curve at point (1,0)

slope =

$
\frac{f(x)-f(a)}{x-a}
$

I'm a little confused on what 'a' equals, but going off my notes im gonna say its 1, being x value in the point

$
\frac{(x-x^5)-(1-1^5)}{x-1}
$

=

$
\frac{x-x^5}{x-1}
$

and thats where I get stuck

factor out the x, you get

$\lim_{x \to 1} \frac {x(1 - x^4)}{x - 1}$

now use the difference of two squares for the numerator
• Sep 19th 2008, 03:41 PM
silencecloak
Quote:

Originally Posted by Jhevon
factor out the x, you get

$\lim_{x \to 1} \frac {x(1 - x^4)}{x - 1}$

now use the difference of two squares for the numerator

$\lim_{x \to 1} \frac { x(1-x)(1+x)(1+x^2)}{x - 1}$

=

$\lim_{x \to 1} \frac{x-(x-1)(1+x)(1+x^2)}{x-1}$

=

$\lim_{x \to 1} \ x-(1+x)(1+x^2)$

=

$\lim_{x \to 1} 1-(1+1)(1+1)$

=

$\lim_{x \to 1} 1-(1+1)(1+1)$

= -3

pretty sure i goofed somewhere or im doing it wrong
• Sep 19th 2008, 03:44 PM
Jhevon
Quote:

Originally Posted by silencecloak
$\lim_{x \to 1} \frac { x(1-x)(1+x)(1+x^2)}{x - 1}$

=

$\lim_{x \to 1} \frac{x{\color{red}-}(x-1)(1+x)(1+x^2)}{x-1}$

well, yeah. obviously if you put stuff there that's not supposed to be there that will happen :p exactly where did that minus sign come from? do you realize that you are changing the value of the function?
• Sep 19th 2008, 03:46 PM
silencecloak
Quote:

Originally Posted by Jhevon
well, yeah. obviously if you put stuff there that's not supposed to be there that will happen :p exactly where did that minus sign come from? do you realize that you are changing the value of the function?

Ya, I was trying to pull a negative out so I could cancel out the denominator

EDIT:

durrr

$\frac{-x+1}{x-1} = -1$

so

$
-x(1+x)(1+x^2) = -4

$
• Sep 19th 2008, 04:16 PM
Jhevon
Quote:

Originally Posted by silencecloak
Ya, I was trying to pull a negative out so I could cancel out the denominator

EDIT:

durrr

$\frac{-x+1}{x-1} = -1$

so

$
-x(1+x)(1+x^2) = -4

$

yes

of course you have to write the limit until you actually take the limit