Curve Sketching

[reardear]

Hello, I'm doing a curve sketching problem and so far I've only dealt with equations, and this is my first encounter with only a graph.
I've gotten results for the first three questions: how am I doing for those? For the fourth question, I guess the constant is kind of throwing me off, but I tried.

Graph of \(\displaystyle y = f(x):\)

(sorry about the quality, I was trying out a phone scanning app and I cropped it, which zoomed it in a lot)

i) Critical Numbers: \(\displaystyle f\prime(x) = 0\) at \(\displaystyle [-1, 3] \:and\: 5\)

ii) Points where \(\displaystyle f\) is not differentiable: \(\displaystyle (3,-2)\) (corner)

iii) Intervals of increase and decrease: Increasing on \(\displaystyle (3,5)\), decreasing on \(\displaystyle (-\infty,-1)\cup(5,\infty)\)

iv) Intervals of concave up and down: Concave up at \(\displaystyle (-\infty,-1)\) and concave down \(\displaystyle (3,\infty)\)

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I have another related problem here, might as well ask since I really don't know how to solve it.

\(\displaystyle f(x) = sin(2x)-cos(2x)\)
\(\displaystyle f\prime(x) = 2cos(2x) + 2sin(2x)\)
\(\displaystyle f\prime\prime(x) = 4cos(2x) - 4sin(2x)\)
\(\displaystyle D = [0,2\pi]\)

i) Intervals of increase and decrease:
Ok so I need the critical numbers, but I don't know how to get \(\displaystyle f\prime(x) = 0\) with the same value \(\displaystyle x\). I can probably figure out the rest after that block.

ii) Intervals of concave up and down:

 

[girodd]

Hello, I'm doing a curve sketching problem and so far I've only dealt with equations, and this is my first encounter with only a graph.
I've gotten results for the first three questions: how am I doing for those? For the fourth question, I guess the constant is kind of throwing me off, but I tried.

Graph of \(\displaystyle y = f(x):\)
View attachment 2399
(sorry about the quality, I was trying out a phone scanning app and I cropped it, which zoomed it in a lot)

i) Critical Numbers: \(\displaystyle f\prime(x) = 0\) at \(\displaystyle -1, 0, 1, 2, 3, 5\)

ii) Points where \(\displaystyle f\) is not differentiable: \(\displaystyle (3,-2)\) (corner)

iii) Intervals of increase and decrease: Increasing on \(\displaystyle (3,5)\), decreasing on \(\displaystyle (-\infty,-1)\cup(5,\infty)\)

iv) Intervals of concave up and down: Not sure.. concave down at \(\displaystyle (-\infty,-1)\) and concave up \(\displaystyle (3,\infty)\)?

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You recognize that the derivative is zero on an entire interval, and this makes every point in the interval a critical value, not just the few that you picked out.
You have the notion of concave up/down reversed in your mind, a smile is concave up and a frown is concave down.


I have another related problem here, might as well ask since I really don't know how to solve it.

\(\displaystyle f(x) = sin(2x)-cos(2x)\)
\(\displaystyle f\prime(x) = 2cos(2x) + 2sin(2x)\)
\(\displaystyle f\prime\prime(x) = 4cos(2x) - 4sin(2x)\)
\(\displaystyle D = [0,2\pi]\)

i) Intervals of increase and decrease:
Ok so I need the critical numbers, but I don't know how to get \(\displaystyle f\prime(x) = 0\) with the same value \(\displaystyle x\). I can probably figure out the rest after that block.

ii) Intervals of concave up and down:[/QUOTE\

2cos(2x)+2sin(2x)=0 if -tan(2x) = 0, do a little algebra to get this.

 

[reardear]

You recognize that the derivative is zero on an entire interval, and this makes every point in the interval a critical value, not just the few that you picked out.
You have the notion of concave up/down reversed in your mind, a smile is concave up and a frown is concave down.

2cos(2x)+2sin(2x)=0 if -tan(2x) = 0, do a little algebra to get this.

Thanks, and yeah - I think I was imagining the wrong graph.

Wouldn't it be
\(\displaystyle 2sin(2x) = -2cos(2x)\)
\(\displaystyle \large \frac{2sin(2x)}{-2cos(2x)} = 1\)
\(\displaystyle -tan(2x) - 1 = 0\)

with that, I got \(\displaystyle \large x = \frac{3\pi}{4},\frac{7\pi}{4}\) to make \(\displaystyle tan(2x) = -1\)
so then, \(\displaystyle -(-1) - 1 = 0\)?

Edit: Nevermind, that doesn't work. \(\displaystyle \large \frac{3\pi}{8}, \frac{7\pi}{8}\) works, but I don't know how to get that

 

[Deleted member 4993]

Thanks, and yeah - I think I was imagining the wrong graph.

Wouldn't it be
\(\displaystyle 2sin(2x) = -2cos(2x)\)
\(\displaystyle \large \frac{2sin(2x)}{-2cos(2x)} = 1\)
\(\displaystyle -tan(2x) - 1 = 0\)

with that, I got \(\displaystyle \large x = \frac{3\pi}{4},\frac{7\pi}{4}\) to make \(\displaystyle tan(2x) = -1\) ..............Incorrect
so then, \(\displaystyle -(-1) - 1 = 0\)?

Edit: Nevermind, that doesn't work. \(\displaystyle \large \frac{3\pi}{8}, \frac{7\pi}{8}\) works, but I don't know how to get that

You had it just a small mistake. It should be:

\(\displaystyle \large 2x = \frac{3\pi}{4},\frac{7\pi}{4}\) to make \(\displaystyle tan(2x) = -1\)

Then

\(\displaystyle \large x = \frac{3\pi}{8},\frac{7\pi}{8}\) to make \(\displaystyle tan(2x) = -1\)

 

[reardear]

You had it just a small mistake. It should be:

\(\displaystyle \large 2x = \frac{3\pi}{4},\frac{7\pi}{4}\) to make \(\displaystyle tan(2x) = -1\)

Then

\(\displaystyle \large x = \frac{3\pi}{8},\frac{7\pi}{8}\) to make \(\displaystyle tan(2x) = -1\)

I see! Thank you.

So now my resulting answers are:
f(x) is increasing on \(\displaystyle (0,\frac{3\pi}{8}) \:\:and\:\: (\frac{7\pi}{8},2\pi)\) on domain \(\displaystyle D = [0,2\pi]\)
f(x) is decreasing on \(\displaystyle (\frac{3\pi}{8},\frac{7\pi}{8})\) on domain \(\displaystyle D = [0,2\pi]\)

Edit: Gonna double check that.. I think there should be more
Edit2: Yup, missed two more critical values on that domain

 

[Deleted member 4993]

I see! Thank you.

So now my resulting answers are:
f(x) is increasing on \(\displaystyle (0,\frac{3\pi}{8}) \:\:and\:\: (\frac{7\pi}{8},2\pi)\) on domain \(\displaystyle D = [0,2\pi]\)
f(x) is decreasing on \(\displaystyle (\frac{3\pi}{8},\frac{7\pi}{8})\) on domain \(\displaystyle D = [0,2\pi]\)

Edit: Gonna double check that.. I think there should be more

Great!!

tan(2x) = -1 → x = 3π/8, 7π/8, 11π/8 & 15π/8 for D[0,2π]

 

[reardear]

Great!!

tan(2x) = -1 → x = 3π/8, 7π/8, 11π/8 & 15π/8 for D[0,2π]

Yeah !!

f(x) is increasing on \(\displaystyle (0,\frac{3\pi}{8}), (\frac{7\pi}{8},\frac{11\pi}{8}) \:and\: (\frac{15\pi}{8}, 2\pi)\) on domain \(\displaystyle D = [0,2\pi]\)
f(x) is decreasing on \(\displaystyle (\frac{3\pi}{8},\frac{7\pi}{8}) \:and\: (\frac{11\pi}{8},\frac{15\pi}{8})\) on domain \(\displaystyle D = [0,2\pi]\)

Annoying to type latex, gonna do it your way now

 

[reardear]

For the concave question I got:
concave up: (0, π/8), (5π/8, 9π/8) and (13π/8, 2π)
concave down: (π/8, 5π/8) and (9π/8, 13π/8)

Also, I was wondering how the first part of my original question (with the graph) is. Is there any way I can show work, or is that pretty much all I can do?