Mixing
Calculus, solving for increasing/decreasing and concavity
Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)
$$f(x)=frac{ln(x)}{x}$$
Find intervals of concavity for the graph of the function
$$f(x)=frac{ln(x)}{x}$$
I have already found the first and second derivative but I am confused on how to solve for it
$$f '(x) = -frac{ln(x)-1}{x^2}$$
$$f ''(x) = frac{2ln(x)-3}{x^3}$$
calculus
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
asked Nov 5 '16 at 22:59
Nick Mazzone
11
add a comment |
Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)
$$f(x)=frac{ln(x)}{x}$$
Find intervals of concavity for the graph of the function
$$f(x)=frac{ln(x)}{x}$$
I have already found the first and second derivative but I am confused on how to solve for it
$$f '(x) = -frac{ln(x)-1}{x^2}$$
$$f ''(x) = frac{2ln(x)-3}{x^3}$$
calculus
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
asked Nov 5 '16 at 22:59
Nick Mazzone
11
add a comment |
Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)
$$f(x)=frac{ln(x)}{x}$$
Find intervals of concavity for the graph of the function
$$f(x)=frac{ln(x)}{x}$$
I have already found the first and second derivative but I am confused on how to solve for it
$$f '(x) = -frac{ln(x)-1}{x^2}$$
$$f ''(x) = frac{2ln(x)-3}{x^3}$$
calculus
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
asked Nov 5 '16 at 22:59
Nick Mazzone
11
Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)
$$f(x)=frac{ln(x)}{x}$$
Find intervals of concavity for the graph of the function
$$f(x)=frac{ln(x)}{x}$$
I have already found the first and second derivative but I am confused on how to solve for it
$$f '(x) = -frac{ln(x)-1}{x^2}$$
$$f ''(x) = frac{2ln(x)-3}{x^3}$$
calculus
calculus
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
asked Nov 5 '16 at 22:59
Nick Mazzone
11
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
asked Nov 5 '16 at 22:59
Nick Mazzone
11
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
edited Nov 9 '17 at 10:55
Teddy38
2,0512520
2,0512520
asked Nov 5 '16 at 22:59
Nick Mazzone
11
asked Nov 5 '16 at 22:59
Nick Mazzone
11
asked Nov 5 '16 at 22:59
Nick Mazzone
11
11
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
So let's first begin with finding out the intervals where the function is increasing and decreasing.
To do this, we would find the critical numbers of the first derivative.
(That is, what values of x, set f'(x) = 0? Solving this, we get:
x= e and x =0
Now that you have your critical numbers, you can form your intervals. This would be: (negative infinity to 0, 0 to e and e to infinity)
You do the same thing for the second derivative to find intervals for concavity
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
Do you really think that $(-infty,0)$ is allowed ?
– Jean Marie
Nov 5 '16 at 23:11
$f$ is not defined for $xle 0.$
– mfl
Nov 5 '16 at 23:11
if f is not defined would that just make the interval (0,e)?
– Nick Mazzone
Nov 5 '16 at 23:13
also thank you all for the help. much appreciated.
– Nick Mazzone
Nov 5 '16 at 23:13
Sorry about that, yup the interval would be (0,e). and (e,infty)
– Siddart Fredrick
Nov 5 '16 at 23:17
|
show 1 more comment
Hint
You have got $$f'(x)=dfrac{1-ln x}{x^2}.$$ If you solve $1-ln x=0$ you get $x=e.$ Note that $x^2$ is positive on $(0,infty)$ which is the domain of $f.$ So, you need to study the sign of $f$ on $(0,e)$ and $(e,infty).$ You get
$$begin{array}{ccc} & (0,e) & (e,infty)\ mathrm{sign}(f') & quad+ & -end{array}$$ Thus, $f$ is increasing on $(0,e)$ and decreasing on $(e,infty).$
Proceed in a similar way with the second derivative. (Note that $x^3$ is positive on $(0,infty)$ and so you only need to study the sign of the numerator.)
answered Nov 5 '16 at 23:07
mfl
26k12141
add a comment |
Intervals of concavity:
$$f''(x)<0 iff ln(x)<3/2 iff ln(x)<ln(e^{3/2}) iff x<e^{3/2}approx 4.48.$$
(the last equivalence being due to the increasing property of $ln$ function).
Thus $begin{cases}0<x<e^{3/2} &Longrightarrow & f text{concave,}\x>e^{3/2} &Longrightarrow & f text{convex.}end{cases}$
One can see/check on the graphics below that point $A$ with abscissa $e^{3/2}$ separates the curve into its concave and convex parts.
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
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
So let's first begin with finding out the intervals where the function is increasing and decreasing.
To do this, we would find the critical numbers of the first derivative.
(That is, what values of x, set f'(x) = 0? Solving this, we get:
x= e and x =0
Now that you have your critical numbers, you can form your intervals. This would be: (negative infinity to 0, 0 to e and e to infinity)
You do the same thing for the second derivative to find intervals for concavity
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
Do you really think that $(-infty,0)$ is allowed ?
– Jean Marie
Nov 5 '16 at 23:11
$f$ is not defined for $xle 0.$
– mfl
Nov 5 '16 at 23:11
if f is not defined would that just make the interval (0,e)?
– Nick Mazzone
Nov 5 '16 at 23:13
also thank you all for the help. much appreciated.
– Nick Mazzone
Nov 5 '16 at 23:13
Sorry about that, yup the interval would be (0,e). and (e,infty)
– Siddart Fredrick
Nov 5 '16 at 23:17
|
show 1 more comment
So let's first begin with finding out the intervals where the function is increasing and decreasing.
To do this, we would find the critical numbers of the first derivative.
(That is, what values of x, set f'(x) = 0? Solving this, we get:
x= e and x =0
Now that you have your critical numbers, you can form your intervals. This would be: (negative infinity to 0, 0 to e and e to infinity)
You do the same thing for the second derivative to find intervals for concavity
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
Do you really think that $(-infty,0)$ is allowed ?
– Jean Marie
Nov 5 '16 at 23:11
$f$ is not defined for $xle 0.$
– mfl
Nov 5 '16 at 23:11
if f is not defined would that just make the interval (0,e)?
– Nick Mazzone
Nov 5 '16 at 23:13
also thank you all for the help. much appreciated.
– Nick Mazzone
Nov 5 '16 at 23:13
Sorry about that, yup the interval would be (0,e). and (e,infty)
– Siddart Fredrick
Nov 5 '16 at 23:17
|
show 1 more comment
So let's first begin with finding out the intervals where the function is increasing and decreasing.
To do this, we would find the critical numbers of the first derivative.
(That is, what values of x, set f'(x) = 0? Solving this, we get:
x= e and x =0
Now that you have your critical numbers, you can form your intervals. This would be: (negative infinity to 0, 0 to e and e to infinity)
You do the same thing for the second derivative to find intervals for concavity
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
So let's first begin with finding out the intervals where the function is increasing and decreasing.
To do this, we would find the critical numbers of the first derivative.
(That is, what values of x, set f'(x) = 0? Solving this, we get:
x= e and x =0
Now that you have your critical numbers, you can form your intervals. This would be: (negative infinity to 0, 0 to e and e to infinity)
You do the same thing for the second derivative to find intervals for concavity
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
answered Nov 5 '16 at 23:07
Siddart Fredrick
31114
31114
Do you really think that $(-infty,0)$ is allowed ?
– Jean Marie
Nov 5 '16 at 23:11
$f$ is not defined for $xle 0.$
– mfl
Nov 5 '16 at 23:11
if f is not defined would that just make the interval (0,e)?
– Nick Mazzone
Nov 5 '16 at 23:13
also thank you all for the help. much appreciated.
– Nick Mazzone
Nov 5 '16 at 23:13
Sorry about that, yup the interval would be (0,e). and (e,infty)
– Siddart Fredrick
Nov 5 '16 at 23:17
|
show 1 more comment
Do you really think that $(-infty,0)$ is allowed ?
– Jean Marie
Nov 5 '16 at 23:11
$f$ is not defined for $xle 0.$
– mfl
Nov 5 '16 at 23:11
if f is not defined would that just make the interval (0,e)?
– Nick Mazzone
Nov 5 '16 at 23:13
also thank you all for the help. much appreciated.
– Nick Mazzone
Nov 5 '16 at 23:13
Sorry about that, yup the interval would be (0,e). and (e,infty)
– Siddart Fredrick
Nov 5 '16 at 23:17
Do you really think that $(-infty,0)$ is allowed ?
– Jean Marie
Nov 5 '16 at 23:11
Do you really think that $(-infty,0)$ is allowed ?
– Jean Marie
Nov 5 '16 at 23:11
$f$ is not defined for $xle 0.$
– mfl
Nov 5 '16 at 23:11
$f$ is not defined for $xle 0.$
– mfl
Nov 5 '16 at 23:11
if f is not defined would that just make the interval (0,e)?
– Nick Mazzone
Nov 5 '16 at 23:13
if f is not defined would that just make the interval (0,e)?
– Nick Mazzone
Nov 5 '16 at 23:13
also thank you all for the help. much appreciated.
– Nick Mazzone
Nov 5 '16 at 23:13
also thank you all for the help. much appreciated.
– Nick Mazzone
Nov 5 '16 at 23:13
Sorry about that, yup the interval would be (0,e). and (e,infty)
– Siddart Fredrick
Nov 5 '16 at 23:17
Sorry about that, yup the interval would be (0,e). and (e,infty)
– Siddart Fredrick
Nov 5 '16 at 23:17
|
show 1 more comment
Hint
You have got $$f'(x)=dfrac{1-ln x}{x^2}.$$ If you solve $1-ln x=0$ you get $x=e.$ Note that $x^2$ is positive on $(0,infty)$ which is the domain of $f.$ So, you need to study the sign of $f$ on $(0,e)$ and $(e,infty).$ You get
$$begin{array}{ccc} & (0,e) & (e,infty)\ mathrm{sign}(f') & quad+ & -end{array}$$ Thus, $f$ is increasing on $(0,e)$ and decreasing on $(e,infty).$
Proceed in a similar way with the second derivative. (Note that $x^3$ is positive on $(0,infty)$ and so you only need to study the sign of the numerator.)
answered Nov 5 '16 at 23:07
mfl
26k12141
add a comment |
Hint
You have got $$f'(x)=dfrac{1-ln x}{x^2}.$$ If you solve $1-ln x=0$ you get $x=e.$ Note that $x^2$ is positive on $(0,infty)$ which is the domain of $f.$ So, you need to study the sign of $f$ on $(0,e)$ and $(e,infty).$ You get
$$begin{array}{ccc} & (0,e) & (e,infty)\ mathrm{sign}(f') & quad+ & -end{array}$$ Thus, $f$ is increasing on $(0,e)$ and decreasing on $(e,infty).$
Proceed in a similar way with the second derivative. (Note that $x^3$ is positive on $(0,infty)$ and so you only need to study the sign of the numerator.)
answered Nov 5 '16 at 23:07
mfl
26k12141
add a comment |
Hint
You have got $$f'(x)=dfrac{1-ln x}{x^2}.$$ If you solve $1-ln x=0$ you get $x=e.$ Note that $x^2$ is positive on $(0,infty)$ which is the domain of $f.$ So, you need to study the sign of $f$ on $(0,e)$ and $(e,infty).$ You get
$$begin{array}{ccc} & (0,e) & (e,infty)\ mathrm{sign}(f') & quad+ & -end{array}$$ Thus, $f$ is increasing on $(0,e)$ and decreasing on $(e,infty).$
Proceed in a similar way with the second derivative. (Note that $x^3$ is positive on $(0,infty)$ and so you only need to study the sign of the numerator.)
answered Nov 5 '16 at 23:07
mfl
26k12141
Hint
You have got $$f'(x)=dfrac{1-ln x}{x^2}.$$ If you solve $1-ln x=0$ you get $x=e.$ Note that $x^2$ is positive on $(0,infty)$ which is the domain of $f.$ So, you need to study the sign of $f$ on $(0,e)$ and $(e,infty).$ You get
$$begin{array}{ccc} & (0,e) & (e,infty)\ mathrm{sign}(f') & quad+ & -end{array}$$ Thus, $f$ is increasing on $(0,e)$ and decreasing on $(e,infty).$
Proceed in a similar way with the second derivative. (Note that $x^3$ is positive on $(0,infty)$ and so you only need to study the sign of the numerator.)
answered Nov 5 '16 at 23:07
mfl
26k12141
answered Nov 5 '16 at 23:07
mfl
26k12141
answered Nov 5 '16 at 23:07
mfl
26k12141
answered Nov 5 '16 at 23:07
mfl
26k12141
26k12141
add a comment |
add a comment |
Intervals of concavity:
$$f''(x)<0 iff ln(x)<3/2 iff ln(x)<ln(e^{3/2}) iff x<e^{3/2}approx 4.48.$$
(the last equivalence being due to the increasing property of $ln$ function).
Thus $begin{cases}0<x<e^{3/2} &Longrightarrow & f text{concave,}\x>e^{3/2} &Longrightarrow & f text{convex.}end{cases}$
One can see/check on the graphics below that point $A$ with abscissa $e^{3/2}$ separates the curve into its concave and convex parts.
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
add a comment |
Intervals of concavity:
$$f''(x)<0 iff ln(x)<3/2 iff ln(x)<ln(e^{3/2}) iff x<e^{3/2}approx 4.48.$$
(the last equivalence being due to the increasing property of $ln$ function).
Thus $begin{cases}0<x<e^{3/2} &Longrightarrow & f text{concave,}\x>e^{3/2} &Longrightarrow & f text{convex.}end{cases}$
One can see/check on the graphics below that point $A$ with abscissa $e^{3/2}$ separates the curve into its concave and convex parts.
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
add a comment |
Intervals of concavity:
$$f''(x)<0 iff ln(x)<3/2 iff ln(x)<ln(e^{3/2}) iff x<e^{3/2}approx 4.48.$$
(the last equivalence being due to the increasing property of $ln$ function).
Thus $begin{cases}0<x<e^{3/2} &Longrightarrow & f text{concave,}\x>e^{3/2} &Longrightarrow & f text{convex.}end{cases}$
One can see/check on the graphics below that point $A$ with abscissa $e^{3/2}$ separates the curve into its concave and convex parts.
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
Intervals of concavity:
$$f''(x)<0 iff ln(x)<3/2 iff ln(x)<ln(e^{3/2}) iff x<e^{3/2}approx 4.48.$$
(the last equivalence being due to the increasing property of $ln$ function).
Thus $begin{cases}0<x<e^{3/2} &Longrightarrow & f text{concave,}\x>e^{3/2} &Longrightarrow & f text{convex.}end{cases}$
One can see/check on the graphics below that point $A$ with abscissa $e^{3/2}$ separates the curve into its concave and convex parts.
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
answered Nov 5 '16 at 23:50
Jean Marie
28.8k41949
28.8k41949
add a comment |
add a comment |
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