How to find the intersection point between 2cosx and x/2











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0
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I'm trying to find the solution to this because I need to find the area between the curves, but I need this intersection point to properly subtract the unnecessary parts.



I know how to do it with polynomials but with 2cosx and x/2 i just don't know what to do.



Thanks






functions definite-integrals area curves







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  • just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
    – Lee
    3 hours ago










  • There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
    – Andreas
    3 hours ago










  • desmos.com/calculator/9ogj6jkpet
    – Mason
    3 hours ago










  • Well if there's no analytic solution (I think thats the only answer) I guess I can use the numeric value for the integrals. Thanks.
    – Cliff
    3 hours ago










  • @Andreas yes, I see now, even after drawing not easy to find the point values
    – Lee
    3 hours ago















up vote
0
down vote

favorite












I'm trying to find the solution to this because I need to find the area between the curves, but I need this intersection point to properly subtract the unnecessary parts.



I know how to do it with polynomials but with 2cosx and x/2 i just don't know what to do.



Thanks






functions definite-integrals area curves







share|cite|improve this question






















  • just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
    – Lee
    3 hours ago










  • There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
    – Andreas
    3 hours ago










  • desmos.com/calculator/9ogj6jkpet
    – Mason
    3 hours ago










  • Well if there's no analytic solution (I think thats the only answer) I guess I can use the numeric value for the integrals. Thanks.
    – Cliff
    3 hours ago










  • @Andreas yes, I see now, even after drawing not easy to find the point values
    – Lee
    3 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm trying to find the solution to this because I need to find the area between the curves, but I need this intersection point to properly subtract the unnecessary parts.



I know how to do it with polynomials but with 2cosx and x/2 i just don't know what to do.



Thanks






functions definite-integrals area curves







share|cite|improve this question













I'm trying to find the solution to this because I need to find the area between the curves, but I need this intersection point to properly subtract the unnecessary parts.



I know how to do it with polynomials but with 2cosx and x/2 i just don't know what to do.



Thanks






functions definite-integrals area curves




functions definite-integrals area curves






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









Cliff

134




134












  • just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
    – Lee
    3 hours ago










  • There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
    – Andreas
    3 hours ago










  • desmos.com/calculator/9ogj6jkpet
    – Mason
    3 hours ago










  • Well if there's no analytic solution (I think thats the only answer) I guess I can use the numeric value for the integrals. Thanks.
    – Cliff
    3 hours ago










  • @Andreas yes, I see now, even after drawing not easy to find the point values
    – Lee
    3 hours ago


















  • just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
    – Lee
    3 hours ago










  • There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
    – Andreas
    3 hours ago










  • desmos.com/calculator/9ogj6jkpet
    – Mason
    3 hours ago










  • Well if there's no analytic solution (I think thats the only answer) I guess I can use the numeric value for the integrals. Thanks.
    – Cliff
    3 hours ago










  • @Andreas yes, I see now, even after drawing not easy to find the point values
    – Lee
    3 hours ago
















just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
3 hours ago




just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
3 hours ago












There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
3 hours ago




There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
3 hours ago












desmos.com/calculator/9ogj6jkpet
– Mason
3 hours ago




desmos.com/calculator/9ogj6jkpet
– Mason
3 hours ago












Well if there's no analytic solution (I think thats the only answer) I guess I can use the numeric value for the integrals. Thanks.
– Cliff
3 hours ago




Well if there's no analytic solution (I think thats the only answer) I guess I can use the numeric value for the integrals. Thanks.
– Cliff
3 hours ago












@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
3 hours ago




@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
3 hours ago










1 Answer
1






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up vote
0
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You could use the fact that both function are continuous and the intermediate value theorem to approximate both intersection points with more "comfortable" numbers. As a start, note the following:



$$
x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
$$
$$
x=0, 2cos x=2 > frac{x}{2}=0
$$
$$
x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
$$






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    1 Answer
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    1 Answer
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    active

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    up vote
    0
    down vote













    You could use the fact that both function are continuous and the intermediate value theorem to approximate both intersection points with more "comfortable" numbers. As a start, note the following:



    $$
    x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
    $$
    $$
    x=0, 2cos x=2 > frac{x}{2}=0
    $$
    $$
    x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      You could use the fact that both function are continuous and the intermediate value theorem to approximate both intersection points with more "comfortable" numbers. As a start, note the following:



      $$
      x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
      $$
      $$
      x=0, 2cos x=2 > frac{x}{2}=0
      $$
      $$
      x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
      $$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        You could use the fact that both function are continuous and the intermediate value theorem to approximate both intersection points with more "comfortable" numbers. As a start, note the following:



        $$
        x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
        $$
        $$
        x=0, 2cos x=2 > frac{x}{2}=0
        $$
        $$
        x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
        $$






        share|cite|improve this answer












        You could use the fact that both function are continuous and the intermediate value theorem to approximate both intersection points with more "comfortable" numbers. As a start, note the following:



        $$
        x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
        $$
        $$
        x=0, 2cos x=2 > frac{x}{2}=0
        $$
        $$
        x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        P.Diddy

        11216




        11216






























             

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