For a function to be continuous at a point c, three conditions must be met:
1) f(c) is defined.
The left graph in my attachment shows an example of a graph where f(c) is NOT defined.
2) lim_{x \rightarrow c} f(x) exists.
The middle graph shows an example of a graph where f(c) is defined, but...
It's called "adding zero." :tongue: I see it more as a trick to eventually introduce a factor that can be canceled. For instance, if the integral was this:
\int \frac{3x}{(x-2)^2} dx
I would subtract and add 6:
= \int \frac{3x - 6 + 6}{(x-2)^2} dx
= \int \frac{3x - 6}{(x-2)^2} dx + \int...
Best I can do is give you some other examples that use the properties that you may need to continue:
4(2+5^{n-3}) = 8 + 4 \cdot 5^{n-3}
-7 \cdot 4^{n-1} = -7 \cdot 4 \cdot 4^{n-2} = -28 \cdot 4^{n-2}
12 \cdot 10^{n-5} - 4 \cdot 10^{n-5} = 8 \cdot 10^{n-5}
Study the above and try your...
5 \cdot 2^{n-2} \ne 10^{n-2}
5 \cdot 3^{n-2} \ne 15^{n-2}
...and so on.
I don't know if this will help, but note that
2^{n-2} = 2\cdot 2^{n-3}
and
3^{n-2} = 3\cdot 3^{n-3}
Isn't the notation wrong? It looks like you want
(f \circ g)(x), (g \circ f)(x) and (f \circ h)(x)
(function composition)
but it looks more like
(fg)(x), (gf)(x) and (fh)(x)
(combining functions by multiplication)
I knew people who were fooled by this sort of problem. It is possible that a local maximum be "lower" than a local minimum. Look at the graphs of secant or cosecant, for example.
(To help visualize what you did earlier, I attached two pics. The red region is what is being rotated around the x-axis. The "Wrong.bmp" file shows what you did, and the "Right.bmp" file shows what the problem is asking.)
The way you had set it up, your heights of the representative rectangle...
So a cousin has asked me for Calculus help, but my Calculus is rusty. She's in Calculus II (of a 3-semester sequence in the US) and is on Work. I decided to make up a problem for her, but I want to make sure I know what I'm doing.
1. Homework Statement
A cylindrical tank (16 feet high with a...
This may be a dumb question, but I'll ask anyway...
1. Homework Statement
Find the values of a and b such that
\lim_{x \rightarrow 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = \sqrt{3}
2. Homework Equations
N/A
3. The Attempt at a Solution
I already have the work and the solution...
Yikes. In some schools all of that should be covered in Algebra 2. In my case, I remember learning conics, logarithms, and exponential functions in Algebra 2. We didn't get to matrices, though (and I didn't see matrices until senior year!).
In the eyes of some schools, however, you did take...
I wouldn't solve it "algebraically." I would solve it by making a two-column table. First, find values of r that would make the left side 0 or undefined (I call these critical values), which appears you have already done. Then in the first column list all critical values and all intervals...
With regards to undergraduate math courses (and beyond) there is no set, linear order. Instead, there are "branches" where different courses fall. The three main branches are:
1) Analysis: Real analysis, complex analysis, ordinary diff. eq., partial diff. eq., harmonic analysis, functional...
First, you should have clarified in the beginning the original problem, because what you wrote:
(tan(A+B)-tanA)/1+tan(A+B)tanA = tanB
looks like this:
\frac{\tan (A+B) - \tan A}{1} + \tan (A+B) \tan A = \tan B
It looks like you changed the right side from "tan A + tan B" to "tan (A+B)". Why...
This may be a big hint, but you will have to rationalize twice. First, rationalize the numerator. Simplify the numerator, but don't multiply it out in the denominator. Then rationalize the denominator. Here, you will need to multiply out two of the factors in the denominator. Eventually...
You can't choose \theta = -\frac{\pi}{4}, because by definition the range of the inverse cotangent function \theta = \cot^{-1} x is 0 < \theta < \pi. There is only one value for \cot^{-1} \left( -1 \right) and that is \frac{3\pi}{4}.