+9466 First let's find the slope of the line 2x + 3y = 21
2x + 3y = 21
Subtract 2x from both sides of the equation.
3y = -2x + 21
Divide through by 3 .
y = - \(\frac23\)x + 7
Now we can see that the slope of this line is - \(\frac23\) . So....
the slope between any two points of a parallel line also = -\(\frac23\)
the slope between (2, -9) and (j, 17) = - \(\frac23\)
\(\frac{17--9}{j-2}\,=\,-\frac23\\~\\ \frac{26}{j-2}\,=\,-\frac23 \\~\\ 26=-\frac23(j-2)\\~\\ -39=j-2\\~\\ -37=j\)
Here's a graph to check this: https://www.desmos.com/calculator/ovina9gch0
+9466 a) g(x) = f(x + 1)
The domain of f is (-1, 1) , so x + 1 has to be in the interval (-1, 1) .
-1 < x + 1 < 1 Subtract 1 from each part of the inequality.
-2 < x < 0
So the domain of g(x) is (-2, 0) .
b) h(x) = f(x) + 1
The domain of f is (-1, 1) , so x has to be in the interval (-1, 1) .
So the domain of g(x) is (-1, 1) .
c) j(x) = f( 1/x )
The domain of f is (-1, 1) , so 1/x has to be in the interval (-1, 1) .
-1 < 1/x < 1
By looking at a graph, here: https://www.desmos.com/calculator/tswnd5ukqx ,
we can see that all x values > 1 cause 1/x to be within the desired range,
and all x values < -1 cause 1/x to be within the desired range.
So the domain of g(x) is (-∞, -1) U (1, ∞) .
d) k(x) = f( √x )
The domain of f is (-1, 1) , so √x has to be in the interval (-1, 1) .
-1 < √x < 1 To solve this inequality, let's split it into two parts.
-1 < √x This is true for all non-negative x values, so...
x ≥ 0
and
√x < 1
x < 1
We can check this with a graph: https://www.desmos.com/calculator/wqoa050mj0
So the domain of k(x) is [0, 1) .
e) \(l(x)=f(\frac{x+1}{x-1})\)
The domain of f is (-1, 1) , so \(\frac{x+1}{x-1}\) has to be in the interval (-1, 1) .
-1 < \(\frac{x+1}{x-1}\) < 1
The easiest way to solve this inequality is, again, with a graph.
So the x values that cause \(\frac{x+1}{x-1}\) to be in the desired range are those < 0 .
So the domain of l(x) is (-∞, 0)
