• Support PF! Buy your school textbooks, materials and every day products Here!

Intersection points

  • Thread starter SeReNiTy
  • Start date
170
0
Hi guys, i'm just wondering is it possible to solve the following using algebra to obtain the points of intersection of the two curves f(x) = 6sqrt(x) and
g(x) = [(x+5)^2]/36

I got to the point where i reconized that the inverse of g(x) = 6sqrt(x) - 5 which looks alot like the function f(x), any hints or solutions to this problem?
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,734
893
Yes, the inverse of g is f(x)- 5. I don't know that that helps a lot in finding points of intersection.

The only way I see of finding the points of intersection is to set f(x)= g(x), square both sides to get rid of the square root, and solve the resulting fourth degree equation
 
TD
Homework Helper
1,020
0
I don't know if you can get there your way (something with the inverse) but tehcnically, it's solvable since you get a 4th-degree polynomial. It won't be 'fun' though. Of course, watch out for introducing solution when squaring.

[tex]\frac{{\left( {x + 5} \right)^2 }}
{{36}} = 6\sqrt x \Leftrightarrow \left( {\frac{{\left( {x + 5} \right)^2 }}
{{36}}} \right)^2 = \left( {6\sqrt x } \right)^2 \Leftrightarrow \frac{{\left( {x + 5} \right)^4 }}
{{1296}} - 36x = 0[/tex]

If you'd want to know, mathematica gives me:

[tex]\begin{gathered}-5 + \frac{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}{2} - \hfill \\
\frac{{\sqrt{\frac{-{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}} -
216\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}} +
\frac{279936}
{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}}
{3}}}}{2}\end{gathered} [/tex]

and

[tex]\begin{gathered}-5 + \frac{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}{2} + \hfill \\
\frac{{\sqrt{\frac{-{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}} -
216\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}} +
\frac{279936}
{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}}
{3}}}}{2}\end{gathered} [/tex]

which are approx: [itex]x \to 0.013452\, \wedge \,x \to 29.150[/itex]
 
Last edited:
170
0
HallsofIvy said:
Yes, the inverse of g is f(x)- 5. I don't know that that helps a lot in finding points of intersection.

The only way I see of finding the points of intersection is to set f(x)= g(x), square both sides to get rid of the square root, and solve the resulting fourth degree equation
Oh, with the inverse thing, i just remembered doing a question a while ago that involved finding the points of intersection between two inverse functions, it was much easier to computer the intersection between one of the functions and y = x since inverse always intersect along that line.

How do i go about solving the 4th degree polynomial?
 
TD
Homework Helper
1,020
0

Related Threads for: Intersection points

Replies
2
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
Top