Mixing
Find the domain and range of the function $f(x) =sqrt{csc(3x)}$
Find the domain of and range of the function f(x) ,and express your
answer in interval form $f(x)=sqrt{csc(3x)}$
I got the domain, like how its restricted and $sin(x)$ can't equal to zero. However, I'm stuck on the range part of the question. I'm not sure how to do it. I got 0 ≤ $sqrt{csc(3x)}$ ≤ 1, which is wrong when I checked online graphing calculator
This is a review question for University Calculus.
trigonometry
edited Sep 20 '17 at 13:34
Abcd
3,02021134
asked Sep 20 '17 at 13:08
Dospalke
22
add a comment |
Find the domain of and range of the function f(x) ,and express your
answer in interval form $f(x)=sqrt{csc(3x)}$
I got the domain, like how its restricted and $sin(x)$ can't equal to zero. However, I'm stuck on the range part of the question. I'm not sure how to do it. I got 0 ≤ $sqrt{csc(3x)}$ ≤ 1, which is wrong when I checked online graphing calculator
This is a review question for University Calculus.
trigonometry
edited Sep 20 '17 at 13:34
Abcd
3,02021134
asked Sep 20 '17 at 13:08
Dospalke
22
hint: $csc(3x)$ goes to infinity when $x$ approaches 0
– Vasya
Sep 20 '17 at 13:13
Welcome to MSE. Please use MathJax.
– José Carlos Santos
Sep 20 '17 at 13:16
you can make a plot of your function!
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:36
see here wolframalpha.com/input/?i=plot+sqrt(csc(3*x))+for+x%3D-2+to+2
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:47
add a comment |
Find the domain of and range of the function f(x) ,and express your
answer in interval form $f(x)=sqrt{csc(3x)}$
I got the domain, like how its restricted and $sin(x)$ can't equal to zero. However, I'm stuck on the range part of the question. I'm not sure how to do it. I got 0 ≤ $sqrt{csc(3x)}$ ≤ 1, which is wrong when I checked online graphing calculator
This is a review question for University Calculus.
trigonometry
edited Sep 20 '17 at 13:34
Abcd
3,02021134
asked Sep 20 '17 at 13:08
Dospalke
22
Find the domain of and range of the function f(x) ,and express your
answer in interval form $f(x)=sqrt{csc(3x)}$
I got the domain, like how its restricted and $sin(x)$ can't equal to zero. However, I'm stuck on the range part of the question. I'm not sure how to do it. I got 0 ≤ $sqrt{csc(3x)}$ ≤ 1, which is wrong when I checked online graphing calculator
This is a review question for University Calculus.
trigonometry
trigonometry
edited Sep 20 '17 at 13:34
Abcd
3,02021134
asked Sep 20 '17 at 13:08
Dospalke
22
edited Sep 20 '17 at 13:34
Abcd
3,02021134
asked Sep 20 '17 at 13:08
Dospalke
22
edited Sep 20 '17 at 13:34
Abcd
3,02021134
edited Sep 20 '17 at 13:34
Abcd
3,02021134
edited Sep 20 '17 at 13:34
Abcd
3,02021134
3,02021134
asked Sep 20 '17 at 13:08
Dospalke
22
asked Sep 20 '17 at 13:08
Dospalke
22
asked Sep 20 '17 at 13:08
Dospalke
22
22
hint: $csc(3x)$ goes to infinity when $x$ approaches 0
– Vasya
Sep 20 '17 at 13:13
Welcome to MSE. Please use MathJax.
– José Carlos Santos
Sep 20 '17 at 13:16
you can make a plot of your function!
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:36
see here wolframalpha.com/input/?i=plot+sqrt(csc(3*x))+for+x%3D-2+to+2
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:47
add a comment |
hint: $csc(3x)$ goes to infinity when $x$ approaches 0
– Vasya
Sep 20 '17 at 13:13
Welcome to MSE. Please use MathJax.
– José Carlos Santos
Sep 20 '17 at 13:16
you can make a plot of your function!
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:36
see here wolframalpha.com/input/?i=plot+sqrt(csc(3*x))+for+x%3D-2+to+2
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:47
hint: $csc(3x)$ goes to infinity when $x$ approaches 0
– Vasya
Sep 20 '17 at 13:13
hint: $csc(3x)$ goes to infinity when $x$ approaches 0
– Vasya
Sep 20 '17 at 13:13
Welcome to MSE. Please use MathJax.
– José Carlos Santos
Sep 20 '17 at 13:16
Welcome to MSE. Please use MathJax.
– José Carlos Santos
Sep 20 '17 at 13:16
you can make a plot of your function!
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:36
you can make a plot of your function!
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:36
see here wolframalpha.com/input/?i=plot+sqrt(csc(3*x))+for+x%3D-2+to+2
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:47
see here wolframalpha.com/input/?i=plot+sqrt(csc(3*x))+for+x%3D-2+to+2
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:47
add a comment |
2 Answers
2
active
oldest
votes
To determine the domain of $sqrt{csc(3x)}$, you have to first look at where $sin(3x) = 0$ and $csc(3x) < 0$. Notice that this is determined in the following manner:
$$sin(3x) =0 implies 3x = sin^{-1}(0) implies 3x = kpi, k in mathbb{Z}, implies x = frac{kpi}{3}, k in mathbb{Z}.$$ We also note that $csc(3x) < 0$ when $sin(3x) < 0$, so we see that $$sin(3x) < 0 implies pi < 3x< 2pi implies frac{pi}{3} + 2kpi < x < frac{2pi}{3} + 2kpi, k in mathbb{Z}.$$ Therefore the domain of $sqrt{csc(3x)}$ is the set of all $x in mathbb{R}$ such that $x neq frac{kpi}{3}$ and $x not in left( frac{pi}{3} + 2kpi, frac{2pi}{3} + 2kpi right)$.
To determine the range of $sqrt{csc(3x)}$ note that $$sqrt{csc(3x)} = frac{1}{sqrt{sin(3x)}}. $$ Since $sin(3x) to 0$ as $x to 0$, the function is clearly unbounded (above).
It is also worth noting that the range of the function $f(g(x))$ is the range of $f$ when restricted to the domain of $g(x)$ such that the function is well defined. Notice that $sqrt{csc(3x)}$ is the composite of $sin(3x)$ and $frac{1}{sqrt{x}}$. Therefore, the range of $sqrt{csc(3x)}$ is the range of $frac{1}{sqrt{x}}$ when restricted to the domain determined above. It so follows that the range of $sqrt{csc(3x)}$ is $[1, infty)$, since the largest value of $sin(3x)$ on the restricted domain is $1$. Hence the minimum of $sqrt{csc(3x)}$ is $sqrt{1}=1$.
Hope this helps.
I disagree with the comments made by Dr. Sonnhard Graubner. One should not be forced to rely on graphical techniques. While the student should be able to graph this type of functions, a purely analytic solutions should be able to be provided.
answered Sep 20 '17 at 13:53
AmorFati
396529
add a comment |
Assuming the function is real-valued:
$f(x)=sqrt{csc(3x)}$
For the expression $sqrt{a}$ to be defined, we need $age0$. Thus, $csc(3x)ge0$.
Since $csc(x)=frac{1}{sin(x)}$, we also need $sin(3x)ne0Leftrightarrow3xne ncdotpiLeftrightarrow xne ncdotfrac{pi}{3}, ninmathbb{Z}$
Now $csc(3x)ge0$ iff $sin(3x)ge0$
We know $sin(x)ge0$ iff $xinleft[2npi;(2n+1)piright], ninmathbb{Z}$
Then $sin(3x)ge0$ iff $xinleft[frac{2n}{3}pi;frac{2n+1}{3}piright], ninmathbb{Z}$
Combining these two restrictions yields the Domain $D={xinmathbb{R}colon exists ninmathbb{Z}left(xinleft(frac{2n}{3}pi;frac{2n+1}{3}piright)right)}$
As seen above, for this domain the expression $sin(3x)$ takes on the values in the interval $(0;1]$
Then $csc(3x)=frac{1}{sin(3x)}$ takes on values in the interval $[1;+infty)$
Finally, this means that $sqrt{csc(3x)}$ will take on values in the interval $[1;+infty)$, meaning that the range is $R={xinmathbb{R}colon xge1}$
Note: Knowing the sines periodic behavior we can see that $f(x)$ behaves the same in any interval $left(frac{2n}{3}pi;frac{2n+1}{3}piright)$ for any $ninmathbb{Z}$ and will take on all the values in its range in any of these intervals.
answered Sep 20 '17 at 14:01
Thorgott
566314
add a comment |
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2 Answers
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2 Answers
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To determine the domain of $sqrt{csc(3x)}$, you have to first look at where $sin(3x) = 0$ and $csc(3x) < 0$. Notice that this is determined in the following manner:
$$sin(3x) =0 implies 3x = sin^{-1}(0) implies 3x = kpi, k in mathbb{Z}, implies x = frac{kpi}{3}, k in mathbb{Z}.$$ We also note that $csc(3x) < 0$ when $sin(3x) < 0$, so we see that $$sin(3x) < 0 implies pi < 3x< 2pi implies frac{pi}{3} + 2kpi < x < frac{2pi}{3} + 2kpi, k in mathbb{Z}.$$ Therefore the domain of $sqrt{csc(3x)}$ is the set of all $x in mathbb{R}$ such that $x neq frac{kpi}{3}$ and $x not in left( frac{pi}{3} + 2kpi, frac{2pi}{3} + 2kpi right)$.
To determine the range of $sqrt{csc(3x)}$ note that $$sqrt{csc(3x)} = frac{1}{sqrt{sin(3x)}}. $$ Since $sin(3x) to 0$ as $x to 0$, the function is clearly unbounded (above).
It is also worth noting that the range of the function $f(g(x))$ is the range of $f$ when restricted to the domain of $g(x)$ such that the function is well defined. Notice that $sqrt{csc(3x)}$ is the composite of $sin(3x)$ and $frac{1}{sqrt{x}}$. Therefore, the range of $sqrt{csc(3x)}$ is the range of $frac{1}{sqrt{x}}$ when restricted to the domain determined above. It so follows that the range of $sqrt{csc(3x)}$ is $[1, infty)$, since the largest value of $sin(3x)$ on the restricted domain is $1$. Hence the minimum of $sqrt{csc(3x)}$ is $sqrt{1}=1$.
Hope this helps.
I disagree with the comments made by Dr. Sonnhard Graubner. One should not be forced to rely on graphical techniques. While the student should be able to graph this type of functions, a purely analytic solutions should be able to be provided.
answered Sep 20 '17 at 13:53
AmorFati
396529
add a comment |
To determine the domain of $sqrt{csc(3x)}$, you have to first look at where $sin(3x) = 0$ and $csc(3x) < 0$. Notice that this is determined in the following manner:
$$sin(3x) =0 implies 3x = sin^{-1}(0) implies 3x = kpi, k in mathbb{Z}, implies x = frac{kpi}{3}, k in mathbb{Z}.$$ We also note that $csc(3x) < 0$ when $sin(3x) < 0$, so we see that $$sin(3x) < 0 implies pi < 3x< 2pi implies frac{pi}{3} + 2kpi < x < frac{2pi}{3} + 2kpi, k in mathbb{Z}.$$ Therefore the domain of $sqrt{csc(3x)}$ is the set of all $x in mathbb{R}$ such that $x neq frac{kpi}{3}$ and $x not in left( frac{pi}{3} + 2kpi, frac{2pi}{3} + 2kpi right)$.
To determine the range of $sqrt{csc(3x)}$ note that $$sqrt{csc(3x)} = frac{1}{sqrt{sin(3x)}}. $$ Since $sin(3x) to 0$ as $x to 0$, the function is clearly unbounded (above).
It is also worth noting that the range of the function $f(g(x))$ is the range of $f$ when restricted to the domain of $g(x)$ such that the function is well defined. Notice that $sqrt{csc(3x)}$ is the composite of $sin(3x)$ and $frac{1}{sqrt{x}}$. Therefore, the range of $sqrt{csc(3x)}$ is the range of $frac{1}{sqrt{x}}$ when restricted to the domain determined above. It so follows that the range of $sqrt{csc(3x)}$ is $[1, infty)$, since the largest value of $sin(3x)$ on the restricted domain is $1$. Hence the minimum of $sqrt{csc(3x)}$ is $sqrt{1}=1$.
Hope this helps.
I disagree with the comments made by Dr. Sonnhard Graubner. One should not be forced to rely on graphical techniques. While the student should be able to graph this type of functions, a purely analytic solutions should be able to be provided.
answered Sep 20 '17 at 13:53
AmorFati
396529
add a comment |
To determine the domain of $sqrt{csc(3x)}$, you have to first look at where $sin(3x) = 0$ and $csc(3x) < 0$. Notice that this is determined in the following manner:
$$sin(3x) =0 implies 3x = sin^{-1}(0) implies 3x = kpi, k in mathbb{Z}, implies x = frac{kpi}{3}, k in mathbb{Z}.$$ We also note that $csc(3x) < 0$ when $sin(3x) < 0$, so we see that $$sin(3x) < 0 implies pi < 3x< 2pi implies frac{pi}{3} + 2kpi < x < frac{2pi}{3} + 2kpi, k in mathbb{Z}.$$ Therefore the domain of $sqrt{csc(3x)}$ is the set of all $x in mathbb{R}$ such that $x neq frac{kpi}{3}$ and $x not in left( frac{pi}{3} + 2kpi, frac{2pi}{3} + 2kpi right)$.
To determine the range of $sqrt{csc(3x)}$ note that $$sqrt{csc(3x)} = frac{1}{sqrt{sin(3x)}}. $$ Since $sin(3x) to 0$ as $x to 0$, the function is clearly unbounded (above).
It is also worth noting that the range of the function $f(g(x))$ is the range of $f$ when restricted to the domain of $g(x)$ such that the function is well defined. Notice that $sqrt{csc(3x)}$ is the composite of $sin(3x)$ and $frac{1}{sqrt{x}}$. Therefore, the range of $sqrt{csc(3x)}$ is the range of $frac{1}{sqrt{x}}$ when restricted to the domain determined above. It so follows that the range of $sqrt{csc(3x)}$ is $[1, infty)$, since the largest value of $sin(3x)$ on the restricted domain is $1$. Hence the minimum of $sqrt{csc(3x)}$ is $sqrt{1}=1$.
Hope this helps.
I disagree with the comments made by Dr. Sonnhard Graubner. One should not be forced to rely on graphical techniques. While the student should be able to graph this type of functions, a purely analytic solutions should be able to be provided.
answered Sep 20 '17 at 13:53
AmorFati
396529
To determine the domain of $sqrt{csc(3x)}$, you have to first look at where $sin(3x) = 0$ and $csc(3x) < 0$. Notice that this is determined in the following manner:
$$sin(3x) =0 implies 3x = sin^{-1}(0) implies 3x = kpi, k in mathbb{Z}, implies x = frac{kpi}{3}, k in mathbb{Z}.$$ We also note that $csc(3x) < 0$ when $sin(3x) < 0$, so we see that $$sin(3x) < 0 implies pi < 3x< 2pi implies frac{pi}{3} + 2kpi < x < frac{2pi}{3} + 2kpi, k in mathbb{Z}.$$ Therefore the domain of $sqrt{csc(3x)}$ is the set of all $x in mathbb{R}$ such that $x neq frac{kpi}{3}$ and $x not in left( frac{pi}{3} + 2kpi, frac{2pi}{3} + 2kpi right)$.
To determine the range of $sqrt{csc(3x)}$ note that $$sqrt{csc(3x)} = frac{1}{sqrt{sin(3x)}}. $$ Since $sin(3x) to 0$ as $x to 0$, the function is clearly unbounded (above).
It is also worth noting that the range of the function $f(g(x))$ is the range of $f$ when restricted to the domain of $g(x)$ such that the function is well defined. Notice that $sqrt{csc(3x)}$ is the composite of $sin(3x)$ and $frac{1}{sqrt{x}}$. Therefore, the range of $sqrt{csc(3x)}$ is the range of $frac{1}{sqrt{x}}$ when restricted to the domain determined above. It so follows that the range of $sqrt{csc(3x)}$ is $[1, infty)$, since the largest value of $sin(3x)$ on the restricted domain is $1$. Hence the minimum of $sqrt{csc(3x)}$ is $sqrt{1}=1$.
Hope this helps.
I disagree with the comments made by Dr. Sonnhard Graubner. One should not be forced to rely on graphical techniques. While the student should be able to graph this type of functions, a purely analytic solutions should be able to be provided.
answered Sep 20 '17 at 13:53
AmorFati
396529
edited Sep 20 '17 at 13:58
answered Sep 20 '17 at 13:53
AmorFati
396529
answered Sep 20 '17 at 13:53
AmorFati
396529
answered Sep 20 '17 at 13:53
AmorFati
396529
396529
add a comment |
add a comment |
Assuming the function is real-valued:
$f(x)=sqrt{csc(3x)}$
For the expression $sqrt{a}$ to be defined, we need $age0$. Thus, $csc(3x)ge0$.
Since $csc(x)=frac{1}{sin(x)}$, we also need $sin(3x)ne0Leftrightarrow3xne ncdotpiLeftrightarrow xne ncdotfrac{pi}{3}, ninmathbb{Z}$
Now $csc(3x)ge0$ iff $sin(3x)ge0$
We know $sin(x)ge0$ iff $xinleft[2npi;(2n+1)piright], ninmathbb{Z}$
Then $sin(3x)ge0$ iff $xinleft[frac{2n}{3}pi;frac{2n+1}{3}piright], ninmathbb{Z}$
Combining these two restrictions yields the Domain $D={xinmathbb{R}colon exists ninmathbb{Z}left(xinleft(frac{2n}{3}pi;frac{2n+1}{3}piright)right)}$
As seen above, for this domain the expression $sin(3x)$ takes on the values in the interval $(0;1]$
Then $csc(3x)=frac{1}{sin(3x)}$ takes on values in the interval $[1;+infty)$
Finally, this means that $sqrt{csc(3x)}$ will take on values in the interval $[1;+infty)$, meaning that the range is $R={xinmathbb{R}colon xge1}$
Note: Knowing the sines periodic behavior we can see that $f(x)$ behaves the same in any interval $left(frac{2n}{3}pi;frac{2n+1}{3}piright)$ for any $ninmathbb{Z}$ and will take on all the values in its range in any of these intervals.
answered Sep 20 '17 at 14:01
Thorgott
566314
add a comment |
Assuming the function is real-valued:
$f(x)=sqrt{csc(3x)}$
For the expression $sqrt{a}$ to be defined, we need $age0$. Thus, $csc(3x)ge0$.
Since $csc(x)=frac{1}{sin(x)}$, we also need $sin(3x)ne0Leftrightarrow3xne ncdotpiLeftrightarrow xne ncdotfrac{pi}{3}, ninmathbb{Z}$
Now $csc(3x)ge0$ iff $sin(3x)ge0$
We know $sin(x)ge0$ iff $xinleft[2npi;(2n+1)piright], ninmathbb{Z}$
Then $sin(3x)ge0$ iff $xinleft[frac{2n}{3}pi;frac{2n+1}{3}piright], ninmathbb{Z}$
Combining these two restrictions yields the Domain $D={xinmathbb{R}colon exists ninmathbb{Z}left(xinleft(frac{2n}{3}pi;frac{2n+1}{3}piright)right)}$
As seen above, for this domain the expression $sin(3x)$ takes on the values in the interval $(0;1]$
Then $csc(3x)=frac{1}{sin(3x)}$ takes on values in the interval $[1;+infty)$
Finally, this means that $sqrt{csc(3x)}$ will take on values in the interval $[1;+infty)$, meaning that the range is $R={xinmathbb{R}colon xge1}$
Note: Knowing the sines periodic behavior we can see that $f(x)$ behaves the same in any interval $left(frac{2n}{3}pi;frac{2n+1}{3}piright)$ for any $ninmathbb{Z}$ and will take on all the values in its range in any of these intervals.
answered Sep 20 '17 at 14:01
Thorgott
566314
add a comment |
Assuming the function is real-valued:
$f(x)=sqrt{csc(3x)}$
For the expression $sqrt{a}$ to be defined, we need $age0$. Thus, $csc(3x)ge0$.
Since $csc(x)=frac{1}{sin(x)}$, we also need $sin(3x)ne0Leftrightarrow3xne ncdotpiLeftrightarrow xne ncdotfrac{pi}{3}, ninmathbb{Z}$
Now $csc(3x)ge0$ iff $sin(3x)ge0$
We know $sin(x)ge0$ iff $xinleft[2npi;(2n+1)piright], ninmathbb{Z}$
Then $sin(3x)ge0$ iff $xinleft[frac{2n}{3}pi;frac{2n+1}{3}piright], ninmathbb{Z}$
Combining these two restrictions yields the Domain $D={xinmathbb{R}colon exists ninmathbb{Z}left(xinleft(frac{2n}{3}pi;frac{2n+1}{3}piright)right)}$
As seen above, for this domain the expression $sin(3x)$ takes on the values in the interval $(0;1]$
Then $csc(3x)=frac{1}{sin(3x)}$ takes on values in the interval $[1;+infty)$
Finally, this means that $sqrt{csc(3x)}$ will take on values in the interval $[1;+infty)$, meaning that the range is $R={xinmathbb{R}colon xge1}$
Note: Knowing the sines periodic behavior we can see that $f(x)$ behaves the same in any interval $left(frac{2n}{3}pi;frac{2n+1}{3}piright)$ for any $ninmathbb{Z}$ and will take on all the values in its range in any of these intervals.
answered Sep 20 '17 at 14:01
Thorgott
566314
Assuming the function is real-valued:
$f(x)=sqrt{csc(3x)}$
For the expression $sqrt{a}$ to be defined, we need $age0$. Thus, $csc(3x)ge0$.
Since $csc(x)=frac{1}{sin(x)}$, we also need $sin(3x)ne0Leftrightarrow3xne ncdotpiLeftrightarrow xne ncdotfrac{pi}{3}, ninmathbb{Z}$
Now $csc(3x)ge0$ iff $sin(3x)ge0$
We know $sin(x)ge0$ iff $xinleft[2npi;(2n+1)piright], ninmathbb{Z}$
Then $sin(3x)ge0$ iff $xinleft[frac{2n}{3}pi;frac{2n+1}{3}piright], ninmathbb{Z}$
Combining these two restrictions yields the Domain $D={xinmathbb{R}colon exists ninmathbb{Z}left(xinleft(frac{2n}{3}pi;frac{2n+1}{3}piright)right)}$
As seen above, for this domain the expression $sin(3x)$ takes on the values in the interval $(0;1]$
Then $csc(3x)=frac{1}{sin(3x)}$ takes on values in the interval $[1;+infty)$
Finally, this means that $sqrt{csc(3x)}$ will take on values in the interval $[1;+infty)$, meaning that the range is $R={xinmathbb{R}colon xge1}$
Note: Knowing the sines periodic behavior we can see that $f(x)$ behaves the same in any interval $left(frac{2n}{3}pi;frac{2n+1}{3}piright)$ for any $ninmathbb{Z}$ and will take on all the values in its range in any of these intervals.
answered Sep 20 '17 at 14:01
Thorgott
566314
answered Sep 20 '17 at 14:01
Thorgott
566314
answered Sep 20 '17 at 14:01
Thorgott
566314
answered Sep 20 '17 at 14:01
Thorgott
566314
566314
add a comment |
add a comment |
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hint: $csc(3x)$ goes to infinity when $x$ approaches 0
– Vasya
Sep 20 '17 at 13:13
Welcome to MSE. Please use MathJax.
– José Carlos Santos
Sep 20 '17 at 13:16
you can make a plot of your function!
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:36
see here wolframalpha.com/input/?i=plot+sqrt(csc(3*x))+for+x%3D-2+to+2
– Dr. Sonnhard Graubner
Sep 20 '17 at 13:47