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
asked Nov 19 at 8:27
Cliff
115
add a comment |
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
asked Nov 19 at 8:27
Cliff
115
just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
Nov 19 at 8:32
There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
Nov 19 at 8:34
desmos.com/calculator/9ogj6jkpet
– Mason
Nov 19 at 8:35
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
Nov 19 at 8:36
@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
Nov 19 at 8:37
add a comment |
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
asked Nov 19 at 8:27
Cliff
115
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
asked Nov 19 at 8:27
Cliff
115
asked Nov 19 at 8:27
Cliff
115
asked Nov 19 at 8:27
Cliff
115
asked Nov 19 at 8:27
Cliff
115
asked Nov 19 at 8:27
Cliff
115
115
just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
Nov 19 at 8:32
There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
Nov 19 at 8:34
desmos.com/calculator/9ogj6jkpet
– Mason
Nov 19 at 8:35
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
Nov 19 at 8:36
@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
Nov 19 at 8:37
add a comment |
just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
Nov 19 at 8:32
There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
Nov 19 at 8:34
desmos.com/calculator/9ogj6jkpet
– Mason
Nov 19 at 8:35
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
Nov 19 at 8:36
@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
Nov 19 at 8:37
just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
Nov 19 at 8:32
just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
Nov 19 at 8:32
There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
Nov 19 at 8:34
There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
Nov 19 at 8:34
desmos.com/calculator/9ogj6jkpet
– Mason
Nov 19 at 8:35
desmos.com/calculator/9ogj6jkpet
– Mason
Nov 19 at 8:35
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
Nov 19 at 8:36
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
Nov 19 at 8:36
@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
Nov 19 at 8:37
@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
Nov 19 at 8:37
add a comment |
1 Answer
1
active
oldest
votes
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:
$$
x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
$$
$$
x=0, 2cos x=2 > frac{x}{2}=0
$$
$$
x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
$$
answered Nov 19 at 8:52
P.Diddy
11216
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:
$$
x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
$$
$$
x=0, 2cos x=2 > frac{x}{2}=0
$$
$$
x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
$$
answered Nov 19 at 8:52
P.Diddy
11216
add a comment |
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:
$$
x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
$$
$$
x=0, 2cos x=2 > frac{x}{2}=0
$$
$$
x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
$$
answered Nov 19 at 8:52
P.Diddy
11216
add a comment |
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:
$$
x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
$$
$$
x=0, 2cos x=2 > frac{x}{2}=0
$$
$$
x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
$$
answered Nov 19 at 8:52
P.Diddy
11216
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:
$$
x=-pi, 2cos x = -2 < frac{x}{2}=-frac{pi}{2}
$$
$$
x=0, 2cos x=2 > frac{x}{2}=0
$$
$$
x=frac{pi}{2}, 2cos x = 0 < frac{x}{2}=frac{pi}{4}
$$
answered Nov 19 at 8:52
P.Diddy
11216
answered Nov 19 at 8:52
P.Diddy
11216
answered Nov 19 at 8:52
P.Diddy
11216
answered Nov 19 at 8:52
P.Diddy
11216
11216
add a comment |
add a comment |
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just draw $f(x)=2cosx$ and $f(x)=x/2$, and you will see the intersection points
– Lee
Nov 19 at 8:32
There is no analytic solution to that problem. Numerically, the intersection is at x≈-2.13333.
– Andreas
Nov 19 at 8:34
desmos.com/calculator/9ogj6jkpet
– Mason
Nov 19 at 8:35
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
Nov 19 at 8:36
@Andreas yes, I see now, even after drawing not easy to find the point values
– Lee
Nov 19 at 8:37