Proving that a line and a function meet in two points
Clash Royale CLAN TAG#URR8PPP
I had an exam today and there was the following question:
Prove that $y=fracx+12$ and $f(x)=frace^2x-12$ meet in two points.
I tried to solve it with the Intermediate value theorem, but without any success. Im not sure that it's the right way.
How should I approach this problem?
calculus algebra-precalculus roots
edited Dec 20 '18 at 21:35
Jam
4,94811431
asked Dec 20 '18 at 19:29
TTaJTa4
734
add a comment |
I had an exam today and there was the following question:
Prove that $y=fracx+12$ and $f(x)=frace^2x-12$ meet in two points.
I tried to solve it with the Intermediate value theorem, but without any success. Im not sure that it's the right way.
How should I approach this problem?
calculus algebra-precalculus roots
edited Dec 20 '18 at 21:35
Jam
4,94811431
asked Dec 20 '18 at 19:29
TTaJTa4
734
add a comment |
I had an exam today and there was the following question:
Prove that $y=fracx+12$ and $f(x)=frace^2x-12$ meet in two points.
I tried to solve it with the Intermediate value theorem, but without any success. Im not sure that it's the right way.
How should I approach this problem?
calculus algebra-precalculus roots
edited Dec 20 '18 at 21:35
Jam
4,94811431
asked Dec 20 '18 at 19:29
TTaJTa4
734
I had an exam today and there was the following question:
Prove that $y=fracx+12$ and $f(x)=frace^2x-12$ meet in two points.
I tried to solve it with the Intermediate value theorem, but without any success. Im not sure that it's the right way.
How should I approach this problem?
calculus algebra-precalculus roots
calculus algebra-precalculus roots
edited Dec 20 '18 at 21:35
Jam
4,94811431
asked Dec 20 '18 at 19:29
TTaJTa4
734
edited Dec 20 '18 at 21:35
Jam
4,94811431
asked Dec 20 '18 at 19:29
TTaJTa4
734
edited Dec 20 '18 at 21:35
Jam
4,94811431
edited Dec 20 '18 at 21:35
Jam
4,94811431
edited Dec 20 '18 at 21:35
Jam
4,94811431
4,94811431
asked Dec 20 '18 at 19:29
TTaJTa4
734
asked Dec 20 '18 at 19:29
TTaJTa4
734
asked Dec 20 '18 at 19:29
TTaJTa4
734
734
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Let $f(x)=(e^2x-1)/2$ and $g(x)=(x+1)/2$.
Let $h(x)=f(x)-g(x)=(e^2x-x-2)/2$. Then,
$$h(-infty)=+infty, h(+infty)=+infty$$
$$h'(x)=(2e^2x-1)/2implies x_0=frac-ln22 is only minima$$
$$h(x_0)=fracln2-34<0$$
Thus, $h(x)$ intersects $x-$axis exactly twice$implies$ $f(x)=g(x)$ exactly twice.
Thank you Rob Arthan for pointing out the mistake in my earlier answer!
answered Dec 20 '18 at 19:36
Ankit Kumar
1
1
You also need do some work with the derivatives to show that there are exactly two points of intersection.
– Rob Arthan
Dec 20 '18 at 20:14
@RobArthan ya. thanks!
– Ankit Kumar
Dec 20 '18 at 21:10
1
+1 My pleasure! Thanks for improving your answer.
– Rob Arthan
Dec 20 '18 at 21:29
add a comment |
hint
let $$g(x)=2(f(x)-y)=e^2x-x-2$$
Observe that
$$g(-2)=e^-4>0$$
$$g(0)=-1<0$$
$$g(2)=e^4-4>0$$
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
This effectively shows that there are $2$ roots but can you extend your argument to showing that there are only $2$ roots?
– Jam
Dec 20 '18 at 21:47
1
@Jam - the way the question is stated, I read it as only requiring a proof that at least 2 roots exist. The intermediate value theorem cannot be used to prove that there are not more. For that, the easiest method is show that the derivative has only one $0$, as Ross has done, and therefore the function can have only 1 turning point. Having a third $0$ would require at least two turning points.
– Paul Sinclair
Dec 21 '18 at 0:09
@PaulSinclair You could be right.
– Jam
Dec 21 '18 at 0:53
add a comment |
And sometimes a graph helps in these cases:
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $f(x)=(e^2x-1)/2$ and $g(x)=(x+1)/2$.
Let $h(x)=f(x)-g(x)=(e^2x-x-2)/2$. Then,
$$h(-infty)=+infty, h(+infty)=+infty$$
$$h'(x)=(2e^2x-1)/2implies x_0=frac-ln22 is only minima$$
$$h(x_0)=fracln2-34<0$$
Thus, $h(x)$ intersects $x-$axis exactly twice$implies$ $f(x)=g(x)$ exactly twice.
Thank you Rob Arthan for pointing out the mistake in my earlier answer!
answered Dec 20 '18 at 19:36
Ankit Kumar
1
1
You also need do some work with the derivatives to show that there are exactly two points of intersection.
– Rob Arthan
Dec 20 '18 at 20:14
@RobArthan ya. thanks!
– Ankit Kumar
Dec 20 '18 at 21:10
1
+1 My pleasure! Thanks for improving your answer.
– Rob Arthan
Dec 20 '18 at 21:29
add a comment |
Let $f(x)=(e^2x-1)/2$ and $g(x)=(x+1)/2$.
Let $h(x)=f(x)-g(x)=(e^2x-x-2)/2$. Then,
$$h(-infty)=+infty, h(+infty)=+infty$$
$$h'(x)=(2e^2x-1)/2implies x_0=frac-ln22 is only minima$$
$$h(x_0)=fracln2-34<0$$
Thus, $h(x)$ intersects $x-$axis exactly twice$implies$ $f(x)=g(x)$ exactly twice.
Thank you Rob Arthan for pointing out the mistake in my earlier answer!
answered Dec 20 '18 at 19:36
Ankit Kumar
1
1
You also need do some work with the derivatives to show that there are exactly two points of intersection.
– Rob Arthan
Dec 20 '18 at 20:14
@RobArthan ya. thanks!
– Ankit Kumar
Dec 20 '18 at 21:10
1
+1 My pleasure! Thanks for improving your answer.
– Rob Arthan
Dec 20 '18 at 21:29
add a comment |
Let $f(x)=(e^2x-1)/2$ and $g(x)=(x+1)/2$.
Let $h(x)=f(x)-g(x)=(e^2x-x-2)/2$. Then,
$$h(-infty)=+infty, h(+infty)=+infty$$
$$h'(x)=(2e^2x-1)/2implies x_0=frac-ln22 is only minima$$
$$h(x_0)=fracln2-34<0$$
Thus, $h(x)$ intersects $x-$axis exactly twice$implies$ $f(x)=g(x)$ exactly twice.
Thank you Rob Arthan for pointing out the mistake in my earlier answer!
answered Dec 20 '18 at 19:36
Ankit Kumar
1
Let $f(x)=(e^2x-1)/2$ and $g(x)=(x+1)/2$.
Let $h(x)=f(x)-g(x)=(e^2x-x-2)/2$. Then,
$$h(-infty)=+infty, h(+infty)=+infty$$
$$h'(x)=(2e^2x-1)/2implies x_0=frac-ln22 is only minima$$
$$h(x_0)=fracln2-34<0$$
Thus, $h(x)$ intersects $x-$axis exactly twice$implies$ $f(x)=g(x)$ exactly twice.
Thank you Rob Arthan for pointing out the mistake in my earlier answer!
answered Dec 20 '18 at 19:36
Ankit Kumar
1
edited Dec 20 '18 at 21:18
answered Dec 20 '18 at 19:36
Ankit Kumar
1
answered Dec 20 '18 at 19:36
Ankit Kumar
1
answered Dec 20 '18 at 19:36
Ankit Kumar
1
1
1
You also need do some work with the derivatives to show that there are exactly two points of intersection.
– Rob Arthan
Dec 20 '18 at 20:14
@RobArthan ya. thanks!
– Ankit Kumar
Dec 20 '18 at 21:10
1
+1 My pleasure! Thanks for improving your answer.
– Rob Arthan
Dec 20 '18 at 21:29
add a comment |
1
You also need do some work with the derivatives to show that there are exactly two points of intersection.
– Rob Arthan
Dec 20 '18 at 20:14
@RobArthan ya. thanks!
– Ankit Kumar
Dec 20 '18 at 21:10
1
+1 My pleasure! Thanks for improving your answer.
– Rob Arthan
Dec 20 '18 at 21:29
1
1
You also need do some work with the derivatives to show that there are exactly two points of intersection.
– Rob Arthan
Dec 20 '18 at 20:14
You also need do some work with the derivatives to show that there are exactly two points of intersection.
– Rob Arthan
Dec 20 '18 at 20:14
@RobArthan ya. thanks!
– Ankit Kumar
Dec 20 '18 at 21:10
@RobArthan ya. thanks!
– Ankit Kumar
Dec 20 '18 at 21:10
1
1
+1 My pleasure! Thanks for improving your answer.
– Rob Arthan
Dec 20 '18 at 21:29
+1 My pleasure! Thanks for improving your answer.
– Rob Arthan
Dec 20 '18 at 21:29
add a comment |
hint
let $$g(x)=2(f(x)-y)=e^2x-x-2$$
Observe that
$$g(-2)=e^-4>0$$
$$g(0)=-1<0$$
$$g(2)=e^4-4>0$$
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
This effectively shows that there are $2$ roots but can you extend your argument to showing that there are only $2$ roots?
– Jam
Dec 20 '18 at 21:47
1
@Jam - the way the question is stated, I read it as only requiring a proof that at least 2 roots exist. The intermediate value theorem cannot be used to prove that there are not more. For that, the easiest method is show that the derivative has only one $0$, as Ross has done, and therefore the function can have only 1 turning point. Having a third $0$ would require at least two turning points.
– Paul Sinclair
Dec 21 '18 at 0:09
@PaulSinclair You could be right.
– Jam
Dec 21 '18 at 0:53
add a comment |
hint
let $$g(x)=2(f(x)-y)=e^2x-x-2$$
Observe that
$$g(-2)=e^-4>0$$
$$g(0)=-1<0$$
$$g(2)=e^4-4>0$$
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
This effectively shows that there are $2$ roots but can you extend your argument to showing that there are only $2$ roots?
– Jam
Dec 20 '18 at 21:47
1
@Jam - the way the question is stated, I read it as only requiring a proof that at least 2 roots exist. The intermediate value theorem cannot be used to prove that there are not more. For that, the easiest method is show that the derivative has only one $0$, as Ross has done, and therefore the function can have only 1 turning point. Having a third $0$ would require at least two turning points.
– Paul Sinclair
Dec 21 '18 at 0:09
@PaulSinclair You could be right.
– Jam
Dec 21 '18 at 0:53
add a comment |
hint
let $$g(x)=2(f(x)-y)=e^2x-x-2$$
Observe that
$$g(-2)=e^-4>0$$
$$g(0)=-1<0$$
$$g(2)=e^4-4>0$$
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
hint
let $$g(x)=2(f(x)-y)=e^2x-x-2$$
Observe that
$$g(-2)=e^-4>0$$
$$g(0)=-1<0$$
$$g(2)=e^4-4>0$$
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
answered Dec 20 '18 at 19:36
hamam_Abdallah
37.9k21634
37.9k21634
This effectively shows that there are $2$ roots but can you extend your argument to showing that there are only $2$ roots?
– Jam
Dec 20 '18 at 21:47
1
@Jam - the way the question is stated, I read it as only requiring a proof that at least 2 roots exist. The intermediate value theorem cannot be used to prove that there are not more. For that, the easiest method is show that the derivative has only one $0$, as Ross has done, and therefore the function can have only 1 turning point. Having a third $0$ would require at least two turning points.
– Paul Sinclair
Dec 21 '18 at 0:09
@PaulSinclair You could be right.
– Jam
Dec 21 '18 at 0:53
add a comment |
This effectively shows that there are $2$ roots but can you extend your argument to showing that there are only $2$ roots?
– Jam
Dec 20 '18 at 21:47
1
@Jam - the way the question is stated, I read it as only requiring a proof that at least 2 roots exist. The intermediate value theorem cannot be used to prove that there are not more. For that, the easiest method is show that the derivative has only one $0$, as Ross has done, and therefore the function can have only 1 turning point. Having a third $0$ would require at least two turning points.
– Paul Sinclair
Dec 21 '18 at 0:09
@PaulSinclair You could be right.
– Jam
Dec 21 '18 at 0:53
This effectively shows that there are $2$ roots but can you extend your argument to showing that there are only $2$ roots?
– Jam
Dec 20 '18 at 21:47
This effectively shows that there are $2$ roots but can you extend your argument to showing that there are only $2$ roots?
– Jam
Dec 20 '18 at 21:47
1
1
@Jam - the way the question is stated, I read it as only requiring a proof that at least 2 roots exist. The intermediate value theorem cannot be used to prove that there are not more. For that, the easiest method is show that the derivative has only one $0$, as Ross has done, and therefore the function can have only 1 turning point. Having a third $0$ would require at least two turning points.
– Paul Sinclair
Dec 21 '18 at 0:09
@Jam - the way the question is stated, I read it as only requiring a proof that at least 2 roots exist. The intermediate value theorem cannot be used to prove that there are not more. For that, the easiest method is show that the derivative has only one $0$, as Ross has done, and therefore the function can have only 1 turning point. Having a third $0$ would require at least two turning points.
– Paul Sinclair
Dec 21 '18 at 0:09
@PaulSinclair You could be right.
– Jam
Dec 21 '18 at 0:53
@PaulSinclair You could be right.
– Jam
Dec 21 '18 at 0:53
add a comment |
And sometimes a graph helps in these cases:
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
add a comment |
And sometimes a graph helps in these cases:
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
add a comment |
And sometimes a graph helps in these cases:
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
And sometimes a graph helps in these cases:
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
answered Dec 20 '18 at 21:40
David G. Stork
9,79521232
9,79521232
add a comment |
add a comment |
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