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...