Mixing
Smallest positive root of $tan(x) - x$
$begingroup$
The smallest positive root of $tan(x)-x=0$ lies in?
Considering $f(x) = tan(x)-x$ to be a function, then differentiaing it, I get $f'(x) = tan^2(x)$. This implies it's an increasing function. So shouldn't $f(x) =0$ for $x > 0$ be impossible because at $x=0$ the function is zero?
calculus
edited Feb 1 at 8:55
Martin R
30.8k33561
asked Feb 1 at 8:21
ScáthachScáthach
1498
$endgroup$
add a comment |
$begingroup$
The smallest positive root of $tan(x)-x=0$ lies in?
Considering $f(x) = tan(x)-x$ to be a function, then differentiaing it, I get $f'(x) = tan^2(x)$. This implies it's an increasing function. So shouldn't $f(x) =0$ for $x > 0$ be impossible because at $x=0$ the function is zero?
calculus
edited Feb 1 at 8:55
Martin R
30.8k33561
asked Feb 1 at 8:21
ScáthachScáthach
1498
$endgroup$
2
$begingroup$
Hint: $f(x)$ is increasing in each interval where it is defined. What happens beyond $frac pi 2$ ?
$endgroup$
– Martin R
Feb 1 at 8:26
$begingroup$
After x=π/2, tanx starts increasing from (-∞ to ∞)and x is increasing like always
$endgroup$
– Scáthach
Feb 1 at 8:30
1
$begingroup$
So it would make sense for them to intersect here...
$endgroup$
– Scáthach
Feb 1 at 8:30
add a comment |
$begingroup$
The smallest positive root of $tan(x)-x=0$ lies in?
Considering $f(x) = tan(x)-x$ to be a function, then differentiaing it, I get $f'(x) = tan^2(x)$. This implies it's an increasing function. So shouldn't $f(x) =0$ for $x > 0$ be impossible because at $x=0$ the function is zero?
calculus
edited Feb 1 at 8:55
Martin R
30.8k33561
asked Feb 1 at 8:21
ScáthachScáthach
1498
$endgroup$
The smallest positive root of $tan(x)-x=0$ lies in?
Considering $f(x) = tan(x)-x$ to be a function, then differentiaing it, I get $f'(x) = tan^2(x)$. This implies it's an increasing function. So shouldn't $f(x) =0$ for $x > 0$ be impossible because at $x=0$ the function is zero?
calculus
calculus
edited Feb 1 at 8:55
Martin R
30.8k33561
asked Feb 1 at 8:21
ScáthachScáthach
1498
edited Feb 1 at 8:55
Martin R
30.8k33561
asked Feb 1 at 8:21
ScáthachScáthach
1498
edited Feb 1 at 8:55
Martin R
30.8k33561
edited Feb 1 at 8:55
Martin R
30.8k33561
edited Feb 1 at 8:55
Martin R
30.8k33561
30.8k33561
asked Feb 1 at 8:21
ScáthachScáthach
1498
asked Feb 1 at 8:21
ScáthachScáthach
1498
asked Feb 1 at 8:21
ScáthachScáthach
1498
1498
2
$begingroup$
Hint: $f(x)$ is increasing in each interval where it is defined. What happens beyond $frac pi 2$ ?
$endgroup$
– Martin R
Feb 1 at 8:26
$begingroup$
After x=π/2, tanx starts increasing from (-∞ to ∞)and x is increasing like always
$endgroup$
– Scáthach
Feb 1 at 8:30
1
$begingroup$
So it would make sense for them to intersect here...
$endgroup$
– Scáthach
Feb 1 at 8:30
add a comment |
2
$begingroup$
Hint: $f(x)$ is increasing in each interval where it is defined. What happens beyond $frac pi 2$ ?
$endgroup$
– Martin R
Feb 1 at 8:26
$begingroup$
After x=π/2, tanx starts increasing from (-∞ to ∞)and x is increasing like always
$endgroup$
– Scáthach
Feb 1 at 8:30
1
$begingroup$
So it would make sense for them to intersect here...
$endgroup$
– Scáthach
Feb 1 at 8:30
2
2
$begingroup$
Hint: $f(x)$ is increasing in each interval where it is defined. What happens beyond $frac pi 2$ ?
$endgroup$
– Martin R
Feb 1 at 8:26
$begingroup$
Hint: $f(x)$ is increasing in each interval where it is defined. What happens beyond $frac pi 2$ ?
$endgroup$
– Martin R
Feb 1 at 8:26
$begingroup$
After x=π/2, tanx starts increasing from (-∞ to ∞)and x is increasing like always
$endgroup$
– Scáthach
Feb 1 at 8:30
$begingroup$
After x=π/2, tanx starts increasing from (-∞ to ∞)and x is increasing like always
$endgroup$
– Scáthach
Feb 1 at 8:30
1
1
$begingroup$
So it would make sense for them to intersect here...
$endgroup$
– Scáthach
Feb 1 at 8:30
$begingroup$
So it would make sense for them to intersect here...
$endgroup$
– Scáthach
Feb 1 at 8:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The conclusion “$f' > 0 implies f$ is increasing” is only valid on intervals where $f$ is defined.
In your case, $f(x) = tan x - x$ is defined on all intervals
$$
I_k = { x in Bbb R, frac{(2k-1)pi}{2} < x < frac{(2k+1)pi}{2} }
$$
with $k in Bbb Z$. Within each interval $I_k$, $f(x)$ is strictly increasing, and approaches $-infty$ at the left boundary, and $+infty$ at the right boundary of the interval.
From this you can conclude that $f(x)$ has (exactly) one root in each $I_k$. As you correctly observed, $x=0$ is the only root in $I_0$. Therefore the smallest positive root is in the interval $I_1 = (pi/2, 3 pi/2)$.
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The conclusion “$f' > 0 implies f$ is increasing” is only valid on intervals where $f$ is defined.
In your case, $f(x) = tan x - x$ is defined on all intervals
$$
I_k = { x in Bbb R, frac{(2k-1)pi}{2} < x < frac{(2k+1)pi}{2} }
$$
with $k in Bbb Z$. Within each interval $I_k$, $f(x)$ is strictly increasing, and approaches $-infty$ at the left boundary, and $+infty$ at the right boundary of the interval.
From this you can conclude that $f(x)$ has (exactly) one root in each $I_k$. As you correctly observed, $x=0$ is the only root in $I_0$. Therefore the smallest positive root is in the interval $I_1 = (pi/2, 3 pi/2)$.
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
$endgroup$
add a comment |
$begingroup$
The conclusion “$f' > 0 implies f$ is increasing” is only valid on intervals where $f$ is defined.
In your case, $f(x) = tan x - x$ is defined on all intervals
$$
I_k = { x in Bbb R, frac{(2k-1)pi}{2} < x < frac{(2k+1)pi}{2} }
$$
with $k in Bbb Z$. Within each interval $I_k$, $f(x)$ is strictly increasing, and approaches $-infty$ at the left boundary, and $+infty$ at the right boundary of the interval.
From this you can conclude that $f(x)$ has (exactly) one root in each $I_k$. As you correctly observed, $x=0$ is the only root in $I_0$. Therefore the smallest positive root is in the interval $I_1 = (pi/2, 3 pi/2)$.
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
$endgroup$
add a comment |
$begingroup$
The conclusion “$f' > 0 implies f$ is increasing” is only valid on intervals where $f$ is defined.
In your case, $f(x) = tan x - x$ is defined on all intervals
$$
I_k = { x in Bbb R, frac{(2k-1)pi}{2} < x < frac{(2k+1)pi}{2} }
$$
with $k in Bbb Z$. Within each interval $I_k$, $f(x)$ is strictly increasing, and approaches $-infty$ at the left boundary, and $+infty$ at the right boundary of the interval.
From this you can conclude that $f(x)$ has (exactly) one root in each $I_k$. As you correctly observed, $x=0$ is the only root in $I_0$. Therefore the smallest positive root is in the interval $I_1 = (pi/2, 3 pi/2)$.
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
$endgroup$
The conclusion “$f' > 0 implies f$ is increasing” is only valid on intervals where $f$ is defined.
In your case, $f(x) = tan x - x$ is defined on all intervals
$$
I_k = { x in Bbb R, frac{(2k-1)pi}{2} < x < frac{(2k+1)pi}{2} }
$$
with $k in Bbb Z$. Within each interval $I_k$, $f(x)$ is strictly increasing, and approaches $-infty$ at the left boundary, and $+infty$ at the right boundary of the interval.
From this you can conclude that $f(x)$ has (exactly) one root in each $I_k$. As you correctly observed, $x=0$ is the only root in $I_0$. Therefore the smallest positive root is in the interval $I_1 = (pi/2, 3 pi/2)$.
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
answered Feb 1 at 8:45
Martin RMartin R
30.8k33561
30.8k33561
add a comment |
add a comment |
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2
$begingroup$
Hint: $f(x)$ is increasing in each interval where it is defined. What happens beyond $frac pi 2$ ?
$endgroup$
– Martin R
Feb 1 at 8:26
$begingroup$
After x=π/2, tanx starts increasing from (-∞ to ∞)and x is increasing like always
$endgroup$
– Scáthach
Feb 1 at 8:30
1
$begingroup$
So it would make sense for them to intersect here...
$endgroup$
– Scáthach
Feb 1 at 8:30