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I need help with continuity and intermediate value theorem
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The problem says:
The roots of $f(x)=x^3-2x-x^2+2$ are $sqrt2$, $-sqrt2$, and $1$. By evaluating $f(-3)$, $f(0)$, $f(1.3)$, and $f(2)$, determine the sign of $f(x)$ on each of the intervals between its roots.
I'm struggling with this subject. It may sound like a stupid excuse but the the book doesn't explain well enough sometimes, I was asked on an exercise to use the method of bisection (which I didn't understand as well as I'd like) and it didn't even mention $fracx_1-x_22$. I had to find that on the internet. Thanks for your time and help.
calculus real-analysis
edited Jul 24 at 2:46
Theo Bendit
12k1843
asked Jul 24 at 2:39
Kevin
1
add a comment |Â
up vote
0
down vote
favorite
The problem says:
The roots of $f(x)=x^3-2x-x^2+2$ are $sqrt2$, $-sqrt2$, and $1$. By evaluating $f(-3)$, $f(0)$, $f(1.3)$, and $f(2)$, determine the sign of $f(x)$ on each of the intervals between its roots.
I'm struggling with this subject. It may sound like a stupid excuse but the the book doesn't explain well enough sometimes, I was asked on an exercise to use the method of bisection (which I didn't understand as well as I'd like) and it didn't even mention $fracx_1-x_22$. I had to find that on the internet. Thanks for your time and help.
calculus real-analysis
edited Jul 24 at 2:46
Theo Bendit
12k1843
asked Jul 24 at 2:39
Kevin
1
2
I'd suggest by starting with drawing the graph of any function that cross the $x$-axis three times. Since it only intersects the $x$-axis three times, that means in between those zeroes, the sign (positive or negative) must be constant.
– Clayton
Jul 24 at 2:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The problem says:
The roots of $f(x)=x^3-2x-x^2+2$ are $sqrt2$, $-sqrt2$, and $1$. By evaluating $f(-3)$, $f(0)$, $f(1.3)$, and $f(2)$, determine the sign of $f(x)$ on each of the intervals between its roots.
I'm struggling with this subject. It may sound like a stupid excuse but the the book doesn't explain well enough sometimes, I was asked on an exercise to use the method of bisection (which I didn't understand as well as I'd like) and it didn't even mention $fracx_1-x_22$. I had to find that on the internet. Thanks for your time and help.
calculus real-analysis
edited Jul 24 at 2:46
Theo Bendit
12k1843
asked Jul 24 at 2:39
Kevin
1
The problem says:
The roots of $f(x)=x^3-2x-x^2+2$ are $sqrt2$, $-sqrt2$, and $1$. By evaluating $f(-3)$, $f(0)$, $f(1.3)$, and $f(2)$, determine the sign of $f(x)$ on each of the intervals between its roots.
I'm struggling with this subject. It may sound like a stupid excuse but the the book doesn't explain well enough sometimes, I was asked on an exercise to use the method of bisection (which I didn't understand as well as I'd like) and it didn't even mention $fracx_1-x_22$. I had to find that on the internet. Thanks for your time and help.
calculus real-analysis
edited Jul 24 at 2:46
Theo Bendit
12k1843
asked Jul 24 at 2:39
Kevin
1
edited Jul 24 at 2:46
Theo Bendit
12k1843
edited Jul 24 at 2:46
Theo Bendit
12k1843
edited Jul 24 at 2:46
Theo Bendit
12k1843
12k1843
asked Jul 24 at 2:39
Kevin
1
asked Jul 24 at 2:39
Kevin
1
asked Jul 24 at 2:39
Kevin
1
1
2
I'd suggest by starting with drawing the graph of any function that cross the $x$-axis three times. Since it only intersects the $x$-axis three times, that means in between those zeroes, the sign (positive or negative) must be constant.
– Clayton
Jul 24 at 2:43
add a comment |Â
2
I'd suggest by starting with drawing the graph of any function that cross the $x$-axis three times. Since it only intersects the $x$-axis three times, that means in between those zeroes, the sign (positive or negative) must be constant.
– Clayton
Jul 24 at 2:43
2
2
I'd suggest by starting with drawing the graph of any function that cross the $x$-axis three times. Since it only intersects the $x$-axis three times, that means in between those zeroes, the sign (positive or negative) must be constant.
– Clayton
Jul 24 at 2:43
I'd suggest by starting with drawing the graph of any function that cross the $x$-axis three times. Since it only intersects the $x$-axis three times, that means in between those zeroes, the sign (positive or negative) must be constant.
– Clayton
Jul 24 at 2:43
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Most people get introduced to the notion of continuity first by the statement "You can draw the graph without lifting pencil off the page". Of course, this bears little resemblance to the limit definition; what they're obliquely referring to is the intermediate value theorem.
Basically it says, if you're drawing the graph of a function, and this graph appears above a horizontal line somewhere, and below the horizontal line somewhere else, then somewhere in between, it has to cross the horizontal line.
In this case, we can use it to see that, between its roots, a continuous function has to stay above or below the $x$-axis, but never change between. Why? Because otherwise the function must cross the $x$-axis (i.e. have a root) between the two roots, which is against assumption.
So, if we look at one point between consecutive roots of a continuous function, its sign (i.e. whether it's positive or negative) will be the same sign as every point on the interval of points between those consecutive roots.
answered Jul 24 at 2:53
Theo Bendit
12k1843
add a comment |Â
up vote
1
down vote
For large $x$ the sign of the polynomial is same as that of $x$ and hence $f(x) >0$ in $(sqrt2,infty)$ and $f(x) <0$ in $(-infty, - sqrt2)$. Next $f(x) $ has same sign as that of $f(0)$ in $(-sqrt2,1)$ ie $f(x) >0$ in this interval. The sign of $f(x) $ in interval $(1,sqrt2)$ can be found out more easily by using value of derivative $f'(1)=-1$. Clearly $f$ is decreasing at $1$ and hence $f(x) <0$ in $(1,sqrt2)$. This is somewhat simpler than using the value of $f(1.3)$ which is difficult to evaluate.
The key here is the intermediate value theorem which implies that between any two consecutive roots the sign of the function remains same.
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
add a comment |Â
up vote
0
down vote
Since f(x) is a cubic polynomial, it is continuous and it has at most 3 real roots. This comes from the fundamental theorem of algebra.
As, it is given that there are 3 real roots, there are exactly 3 roots, and it is not possible for the curve to cross the x-axis between $-sqrt 2$ and $1$
So, if we examine $f(x)$ at any $x$ in the interval $(-sqrt 2, 1), f(x)$ must be on that side of $0$ for all $x$ in the interval. $x = 0$ is inside the interval and $f(0) = 2$ so $f(x) > 0$ for all $x in (-sqrt 2, 1).$ Use this logic to evaluate the other intervals. $(-infty, -sqrt 2),(1,sqrt 2)$ and $(sqrt 2, infty)$
answered Jul 24 at 3:20
Doug M
39.1k31749
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Most people get introduced to the notion of continuity first by the statement "You can draw the graph without lifting pencil off the page". Of course, this bears little resemblance to the limit definition; what they're obliquely referring to is the intermediate value theorem.
Basically it says, if you're drawing the graph of a function, and this graph appears above a horizontal line somewhere, and below the horizontal line somewhere else, then somewhere in between, it has to cross the horizontal line.
In this case, we can use it to see that, between its roots, a continuous function has to stay above or below the $x$-axis, but never change between. Why? Because otherwise the function must cross the $x$-axis (i.e. have a root) between the two roots, which is against assumption.
So, if we look at one point between consecutive roots of a continuous function, its sign (i.e. whether it's positive or negative) will be the same sign as every point on the interval of points between those consecutive roots.
answered Jul 24 at 2:53
Theo Bendit
12k1843
add a comment |Â
up vote
2
down vote
Most people get introduced to the notion of continuity first by the statement "You can draw the graph without lifting pencil off the page". Of course, this bears little resemblance to the limit definition; what they're obliquely referring to is the intermediate value theorem.
Basically it says, if you're drawing the graph of a function, and this graph appears above a horizontal line somewhere, and below the horizontal line somewhere else, then somewhere in between, it has to cross the horizontal line.
In this case, we can use it to see that, between its roots, a continuous function has to stay above or below the $x$-axis, but never change between. Why? Because otherwise the function must cross the $x$-axis (i.e. have a root) between the two roots, which is against assumption.
So, if we look at one point between consecutive roots of a continuous function, its sign (i.e. whether it's positive or negative) will be the same sign as every point on the interval of points between those consecutive roots.
answered Jul 24 at 2:53
Theo Bendit
12k1843
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Most people get introduced to the notion of continuity first by the statement "You can draw the graph without lifting pencil off the page". Of course, this bears little resemblance to the limit definition; what they're obliquely referring to is the intermediate value theorem.
Basically it says, if you're drawing the graph of a function, and this graph appears above a horizontal line somewhere, and below the horizontal line somewhere else, then somewhere in between, it has to cross the horizontal line.
In this case, we can use it to see that, between its roots, a continuous function has to stay above or below the $x$-axis, but never change between. Why? Because otherwise the function must cross the $x$-axis (i.e. have a root) between the two roots, which is against assumption.
So, if we look at one point between consecutive roots of a continuous function, its sign (i.e. whether it's positive or negative) will be the same sign as every point on the interval of points between those consecutive roots.
answered Jul 24 at 2:53
Theo Bendit
12k1843
Most people get introduced to the notion of continuity first by the statement "You can draw the graph without lifting pencil off the page". Of course, this bears little resemblance to the limit definition; what they're obliquely referring to is the intermediate value theorem.
Basically it says, if you're drawing the graph of a function, and this graph appears above a horizontal line somewhere, and below the horizontal line somewhere else, then somewhere in between, it has to cross the horizontal line.
In this case, we can use it to see that, between its roots, a continuous function has to stay above or below the $x$-axis, but never change between. Why? Because otherwise the function must cross the $x$-axis (i.e. have a root) between the two roots, which is against assumption.
So, if we look at one point between consecutive roots of a continuous function, its sign (i.e. whether it's positive or negative) will be the same sign as every point on the interval of points between those consecutive roots.
answered Jul 24 at 2:53
Theo Bendit
12k1843
answered Jul 24 at 2:53
Theo Bendit
12k1843
answered Jul 24 at 2:53
Theo Bendit
12k1843
answered Jul 24 at 2:53
Theo Bendit
12k1843
12k1843
add a comment |Â
add a comment |Â
up vote
1
down vote
For large $x$ the sign of the polynomial is same as that of $x$ and hence $f(x) >0$ in $(sqrt2,infty)$ and $f(x) <0$ in $(-infty, - sqrt2)$. Next $f(x) $ has same sign as that of $f(0)$ in $(-sqrt2,1)$ ie $f(x) >0$ in this interval. The sign of $f(x) $ in interval $(1,sqrt2)$ can be found out more easily by using value of derivative $f'(1)=-1$. Clearly $f$ is decreasing at $1$ and hence $f(x) <0$ in $(1,sqrt2)$. This is somewhat simpler than using the value of $f(1.3)$ which is difficult to evaluate.
The key here is the intermediate value theorem which implies that between any two consecutive roots the sign of the function remains same.
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
add a comment |Â
up vote
1
down vote
For large $x$ the sign of the polynomial is same as that of $x$ and hence $f(x) >0$ in $(sqrt2,infty)$ and $f(x) <0$ in $(-infty, - sqrt2)$. Next $f(x) $ has same sign as that of $f(0)$ in $(-sqrt2,1)$ ie $f(x) >0$ in this interval. The sign of $f(x) $ in interval $(1,sqrt2)$ can be found out more easily by using value of derivative $f'(1)=-1$. Clearly $f$ is decreasing at $1$ and hence $f(x) <0$ in $(1,sqrt2)$. This is somewhat simpler than using the value of $f(1.3)$ which is difficult to evaluate.
The key here is the intermediate value theorem which implies that between any two consecutive roots the sign of the function remains same.
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For large $x$ the sign of the polynomial is same as that of $x$ and hence $f(x) >0$ in $(sqrt2,infty)$ and $f(x) <0$ in $(-infty, - sqrt2)$. Next $f(x) $ has same sign as that of $f(0)$ in $(-sqrt2,1)$ ie $f(x) >0$ in this interval. The sign of $f(x) $ in interval $(1,sqrt2)$ can be found out more easily by using value of derivative $f'(1)=-1$. Clearly $f$ is decreasing at $1$ and hence $f(x) <0$ in $(1,sqrt2)$. This is somewhat simpler than using the value of $f(1.3)$ which is difficult to evaluate.
The key here is the intermediate value theorem which implies that between any two consecutive roots the sign of the function remains same.
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
For large $x$ the sign of the polynomial is same as that of $x$ and hence $f(x) >0$ in $(sqrt2,infty)$ and $f(x) <0$ in $(-infty, - sqrt2)$. Next $f(x) $ has same sign as that of $f(0)$ in $(-sqrt2,1)$ ie $f(x) >0$ in this interval. The sign of $f(x) $ in interval $(1,sqrt2)$ can be found out more easily by using value of derivative $f'(1)=-1$. Clearly $f$ is decreasing at $1$ and hence $f(x) <0$ in $(1,sqrt2)$. This is somewhat simpler than using the value of $f(1.3)$ which is difficult to evaluate.
The key here is the intermediate value theorem which implies that between any two consecutive roots the sign of the function remains same.
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
answered Jul 24 at 3:47
Paramanand Singh
45.1k553142
45.1k553142
add a comment |Â
add a comment |Â
up vote
0
down vote
Since f(x) is a cubic polynomial, it is continuous and it has at most 3 real roots. This comes from the fundamental theorem of algebra.
As, it is given that there are 3 real roots, there are exactly 3 roots, and it is not possible for the curve to cross the x-axis between $-sqrt 2$ and $1$
So, if we examine $f(x)$ at any $x$ in the interval $(-sqrt 2, 1), f(x)$ must be on that side of $0$ for all $x$ in the interval. $x = 0$ is inside the interval and $f(0) = 2$ so $f(x) > 0$ for all $x in (-sqrt 2, 1).$ Use this logic to evaluate the other intervals. $(-infty, -sqrt 2),(1,sqrt 2)$ and $(sqrt 2, infty)$
answered Jul 24 at 3:20
Doug M
39.1k31749
add a comment |Â
up vote
0
down vote
Since f(x) is a cubic polynomial, it is continuous and it has at most 3 real roots. This comes from the fundamental theorem of algebra.
As, it is given that there are 3 real roots, there are exactly 3 roots, and it is not possible for the curve to cross the x-axis between $-sqrt 2$ and $1$
So, if we examine $f(x)$ at any $x$ in the interval $(-sqrt 2, 1), f(x)$ must be on that side of $0$ for all $x$ in the interval. $x = 0$ is inside the interval and $f(0) = 2$ so $f(x) > 0$ for all $x in (-sqrt 2, 1).$ Use this logic to evaluate the other intervals. $(-infty, -sqrt 2),(1,sqrt 2)$ and $(sqrt 2, infty)$
answered Jul 24 at 3:20
Doug M
39.1k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since f(x) is a cubic polynomial, it is continuous and it has at most 3 real roots. This comes from the fundamental theorem of algebra.
As, it is given that there are 3 real roots, there are exactly 3 roots, and it is not possible for the curve to cross the x-axis between $-sqrt 2$ and $1$
So, if we examine $f(x)$ at any $x$ in the interval $(-sqrt 2, 1), f(x)$ must be on that side of $0$ for all $x$ in the interval. $x = 0$ is inside the interval and $f(0) = 2$ so $f(x) > 0$ for all $x in (-sqrt 2, 1).$ Use this logic to evaluate the other intervals. $(-infty, -sqrt 2),(1,sqrt 2)$ and $(sqrt 2, infty)$
answered Jul 24 at 3:20
Doug M
39.1k31749
Since f(x) is a cubic polynomial, it is continuous and it has at most 3 real roots. This comes from the fundamental theorem of algebra.
As, it is given that there are 3 real roots, there are exactly 3 roots, and it is not possible for the curve to cross the x-axis between $-sqrt 2$ and $1$
So, if we examine $f(x)$ at any $x$ in the interval $(-sqrt 2, 1), f(x)$ must be on that side of $0$ for all $x$ in the interval. $x = 0$ is inside the interval and $f(0) = 2$ so $f(x) > 0$ for all $x in (-sqrt 2, 1).$ Use this logic to evaluate the other intervals. $(-infty, -sqrt 2),(1,sqrt 2)$ and $(sqrt 2, infty)$
answered Jul 24 at 3:20
Doug M
39.1k31749
answered Jul 24 at 3:20
Doug M
39.1k31749
answered Jul 24 at 3:20
Doug M
39.1k31749
answered Jul 24 at 3:20
Doug M
39.1k31749
39.1k31749
add a comment |Â
add a comment |Â
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2
I'd suggest by starting with drawing the graph of any function that cross the $x$-axis three times. Since it only intersects the $x$-axis three times, that means in between those zeroes, the sign (positive or negative) must be constant.
– Clayton
Jul 24 at 2:43