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Today, 19:21
domain is $x \in \mathbb{R}, \, x \ne 0$
range is $y \in \mathbb{R}, \, y \ne 0$
why so difficult to interpret the graph?
y=\frac{1}{x}
1 replies  22 view(s)

Today, 19:08
$y = \dfrac{1}{x}$ is a basic parent function whose graph is easily sketched ... so, sketch it and answer your own question.
1 replies  18 view(s)

Today, 19:02
Interesting article!
I work with large amounts of data every day and would consider myself an expert in the field when considering the whole...
1 replies  21 view(s)

Today, 18:01
no.
the denominator on the left side of the original inequality is $\sqrt{x}$, which has to be strictly greater than zero $\implies x > 0$
once...
11 replies  109 view(s)

Today, 17:43
Finding the inverse of a function is not always easy.
Sometimes the domain of the function in question must be restricted to force a 11...
3 replies  34 view(s)

Today, 17:32
if you can find the inverse ... sometimes easier said than done
3 replies  38 view(s)

Today, 16:57
Really excited about this blog post by yours truly. Wineman Technology is going places!
1 replies  21 view(s)

Today, 15:28
An alternate approach would be to factor the LHS as the sum of two cubes:
...
6 replies  66 view(s)

Today, 13:34
Here is this week's POTW:

Let $S_0$ be a finite set of positive integers. We define finite sets $S_1,S_2,\ldots$ of positive integers as...
0 replies  23 view(s)

Today, 13:33
No one answered this week's question, the solution of which follows:
For $t$ real and not a multiple of $\pi$, write $g(t) =\frac{f(\cos t)}{\sin...
1 replies  73 view(s)

Yesterday, 19:56
you look at the values of $y$ ... $y \in (\infty,\infty)$
y=12512x^3;
3 replies  38 view(s)

Yesterday, 19:52
$f(x)=12512x$, a 11 function
inverse is $f^{1}(x) = \dfrac{125x}{12}$
the domain of $f^{1}(x)$ is $\mathbb{R} \implies$ range of $f(x)$ is...
3 replies  34 view(s)

Yesterday, 18:45
I tested intervals ... look again at the original inequality. What is the first thing you can say about the domain of $x$?
11 replies  109 view(s)

Yesterday, 17:04
$r_A = (15\sqrt{3} \cdot t)i + (15 \cdot t)j$
$r_B = (20+v\cos{\theta} \cdot t)i + (v\sin{\theta} \cdot t) j$
$v\sin{\theta} = 15 \implies...
5 replies  72 view(s)

Yesterday, 16:10
No, but either way will work, because of the commutative property of addition.. :D
6 replies  74 view(s)

Yesterday, 15:45
In the formula you've used, we have $a=1$...and so after backsubstituting for $u$:
I=\arctan\left(\sqrt{x^2+4x+3}\right)+C
6 replies  74 view(s)

Yesterday, 14:51
As long as the radical is not a factor in a denominator (in that case the radicand cannot be zero), then yes that rule can be applied for domain...
17 replies  217 view(s)

Yesterday, 14:42
no. should look like this ...
0((b/a)>
why?
11 replies  109 view(s)

Yesterday, 14:15
Even though PreCalculus is taught in universities, it's really considered a remedial course at that level, and so that's why we classify it as a...
17 replies  217 view(s)
 Date of Birth
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 BS Mathematics, East Tennessee State University
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