# point on tangent

Printable View

• Aug 6th 2009, 04:08 PM
skeske1234
point on tangent
Show that the tangent to the curve y=(x^2+x-2)^3+3 at the point (1,3) is also the tangent to the curve at another point.

This is what I did so far...
I found the derivative of the curve and I got:
y'=6x^5+15x^4-16x^3-36x^2+16x+12
then I found
f'(1)=-3
and then I put
-3=6x^5+15x^4-16x^3-36x^2+16x+12
and now I am trying to solve for x, BUT I cannot find any factors of x... so? am I doing it right?
• Aug 6th 2009, 04:58 PM
eXist
Well when I take the derivative (Using the chain rule) I get:

\$\displaystyle y' = 3(x^2 + x - 2)^2(2x + 1)\$

and when I find the slope at 1:

\$\displaystyle y'(1) = 3(1^2 + 1 - 2)^2(2(1) + 1) = 3(0)^2(3) = 0\$

So, I'm finding 0 as the slope of the tangent line at the point (1, 3).

Then you have to ask yourself what values of x make this true:

\$\displaystyle 0 = 3(x^2 + x -2)^2(2x + 1)\$

We found one solution, that's when \$\displaystyle x = 1\$, so find the other, it's already factored.

Good Luck - Chad
• Aug 6th 2009, 05:24 PM
ncepeda
Quote:

Originally Posted by skeske1234
Show that the tangent to the curve y=(x^2+x-2)^3+3 at the point (1,3) is also the tangent to the curve at another point.

This is what I did so far...
I found the derivative of the curve and I got:
y'=6x^5+15x^4-16x^3-36x^2+16x+12
then I found
f'(1)=-3
and then I put
-3=6x^5+15x^4-16x^3-36x^2+16x+12
and now I am trying to solve for x, BUT I cannot find any factors of x... so? am I doing it right?

Wow, if you did this by hand this must have took a century to work it all out. Check your derivative I think you may have an error somewhere with all those positive and negative and combining like terms may have messed something up. Let's take an easier approach at the derivative of this thing.

y = (x^2+x-2)^3 + 3

If you recall the chain rule. The rule can be found in the link below as well as some examples.
(The Chain Rule)

3(x^2+x-2)^2 * (2x+1) + 0

y'= 3(x^2+x-2)(2x+1)

Please post if you have more questions.