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• Yesterday, 16:32
Hi FallArk, The function in part (c) is not differentiable at $x = c$, for it is discontinuous at that point. Just check the one-sided limits. ...
3 replies | 69 view(s)
• Yesterday, 12:24
Just to follow up, here is the completed table: Sum $S$ Probability of $S$: $P(S)$ Net Gain/Loss (in dollars) $G$ Product $G\cdot P(S)$
2 replies | 79 view(s)
• March 26th, 2017, 22:53
Euge replied to a thread [SOLVED] Real Analysis in Analysis
No. Use the conditions $f \ge 0$ and $\int_a^b f = 0$, along with basic properties of the integral to prove that result.
12 replies | 181 view(s)
• March 26th, 2017, 22:32
That is the radius of the circle after it has been increased by $a$ units. If I tell you that my weight increased by 20 lbs., then you know my...
4 replies | 84 view(s)
• March 26th, 2017, 21:35
Let's let $0<a$ be the number units the radius must be increased. And so the change in area we can write as: \Delta A=\pi(r+a)^2-\pi r^2=b Now...
4 replies | 84 view(s)
• March 26th, 2017, 21:25
Euge replied to a thread Ball Application in Pre-Calculus
Yes. Yes.
2 replies | 40 view(s)
• March 26th, 2017, 21:19
Euge replied to a thread Find All Real Solutions in Pre-Calculus
Yes, it's correct.
2 replies | 55 view(s)
• March 26th, 2017, 21:15
Euge replied to a thread [SOLVED] Real Analysis in Analysis
Almost. You have a typo in the last line: the fraction $\frac{x + h - h}{h}$ should be $$\frac{x + h - \color{red}{x}}{h}$$
12 replies | 181 view(s)
• March 26th, 2017, 12:39
Euge replied to a thread [SOLVED] Real Analysis in Analysis
To prove problem 1 directly, fix $x\in$ and show that $$f(x) = \lim_{h\to 0^+} \frac{1}{h}\int_x^{x+h} f(t)\, dt$$ Then show that for all...
12 replies | 181 view(s)
• March 26th, 2017, 11:46
Rido12 replied to a thread copper(2) oxide in Other Topics
There are many exceptions. I remember back in my first year chemistry course, my professor criticized the textbook for providing incorrect...
5 replies | 78 view(s)
• March 26th, 2017, 02:05
Euge replied to a thread [SOLVED] Real Analysis in Analysis
Well, the difference between the Riemann sum and integral is made less than $\epsilon$ in magnitude when $n\ge N$, but since $\epsilon$ was...
12 replies | 181 view(s)
• March 26th, 2017, 01:53
Euge replied to a thread [SOLVED] Real Analysis in Analysis
Not quite. Fix $k$. If $n \ge N$ and $x\in \left$, then $\lvert \frac{k}{n} - x\rvert \le \frac{1}{n} \le \frac{1}{N} < \delta$. Thus, for all $n \ge... 12 replies | 181 view(s) • March 26th, 2017, 00:35 Euge replied to a thread [SOLVED] Real Analysis in Analysis Hi joypav, You're not bothering us with your questions, so feel free to ask whenever you have trouble. :) 12 replies | 181 view(s) • March 25th, 2017, 23:17 Yes. Let$h > 0$such that$c + h\in (a,b)$. Then $$F(c + h) - F(c) - f(c+)h = \int_c^{c+h} \, dt$$ Let$\epsilon > 0$. There exists a$\delta...
7 replies | 173 view(s)
• March 25th, 2017, 21:53
MarkFL replied to a thread The Distance Across in Geometry
\overline{MK}=\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2}=\sqrt{2\left(a^2+b^2\right)}=\sqrt{2(50)}=\sqrt{100}=10 :D
8 replies | 91 view(s)
• March 25th, 2017, 21:07
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
I arbitrarily chose another point on the constraint, so that we could do a comparison like I mentioned just now in the other thread. :D
9 replies | 90 view(s)
• March 25th, 2017, 21:04
MarkFL replied to a thread Lagrange Multipliers in Calculus
I chose the point as it is on the constraint. Using that point, we can determine if our one critical point is a maximum or a minimum. If the...
9 replies | 75 view(s)
• March 25th, 2017, 19:21
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
I agree that the point $(2,2)$, is the only one that meets all criteria. Now we need to compare the value of $f$ at another point on the constraint,...
9 replies | 90 view(s)
• March 25th, 2017, 19:08
MarkFL replied to a thread Lagrange Multipliers in Calculus
I agree that of the 3 critical points, $(1,1)$ is the only one in quadrant I. Now, we know this is either a maximum or a minimum, and to determine...
9 replies | 75 view(s)
• March 25th, 2017, 11:15
MarkFL replied to a thread Factoring...6 in Pre-Calculus
It might be more clear to state something like the following: The difference of cubes formula states: p^3-q^3=(p-q)\left(p^2+pq+q^2\right) ...
5 replies | 77 view(s)
• March 25th, 2017, 10:47
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
Consider: e^u=0 What do you get when solving for $u$? Okay, you correctly found $x^2=y^2$...what do you get when you substitute for...
9 replies | 90 view(s)
• March 25th, 2017, 10:39
MarkFL replied to a thread Lagrange Multipliers in Calculus
What I would do is use the constraint to determine $y=2-x$. Now substitute for $y$ in both equations you mentioned, and solve for $x$, then your...
9 replies | 75 view(s)
• March 25th, 2017, 02:30
This is a calculus question...please don't continue to post calculus questions in other forums. If given: ...
3 replies | 87 view(s)
• March 24th, 2017, 23:12
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
If you solve both equations for $\lambda$ and then equate the results, you obtain: \frac{ye^{xy}}{2x}=\frac{xe^{xy}}{2y} Multiply through by 2:...
9 replies | 90 view(s)
• March 24th, 2017, 21:24
MarkFL replied to a thread Lagrange Multipliers in Calculus
Okay, so what this implies is: \frac{x}{\sqrt{6-x^2-y^2}}=\frac{y}{\sqrt{6-x^2-y^2}} Cross-multiply: x\sqrt{6-x^2-y^2}=y\sqrt{6-x^2-y^2} ...
9 replies | 75 view(s)
• March 24th, 2017, 19:07
No, but one argues that a monotone function on a closed interval  belongs to $R$. So then $F$ would make sense. Use an $\epsilon-\delta$ argument...
7 replies | 173 view(s)
• March 24th, 2017, 16:18
MarkFL replied to a thread The Distance Across in Geometry
Using the Pythagorean theorem, we find: \overline{MK}=\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2}=\sqrt{2\left(a^2+b^2\right)} Now, we know that...
8 replies | 91 view(s)
• March 24th, 2017, 06:10
MarkFL replied to a thread [SOLVED] Minimum of function under constraint in Calculus
I would use W|A: W|A - optimize 2x+y subject to xy=18
8 replies | 262 view(s)
• March 24th, 2017, 04:29
The objective function is linear, so it describes a plane, and so I don't believe there will be any saddle points, or in fact any critical points...
4 replies | 63 view(s)
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