- #1
sportsguy3675
- 45
- 0
Little help please and some work checking.
Given [tex]f(x)=3x+2[/tex] and [tex]g(x)=\frac{x-4}{2x}[/tex]
It asks for [tex]g(\frac{1}{x})[/tex]
So substitute in: [tex]\frac{\frac{1}{x} - 4}{2(\frac{1}{x})}[/tex]
Simplify: [tex]\frac{\frac{1}{x} - 4}{\frac{2}{x}}[/tex]
I multiply top and bottom by x right?
That would give: [tex]\frac{1 - 4x}{2}[/tex] and [tex]D_{f} = \Re[/tex] Correct?
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Then on this other problem I have a graph of different functions f and g. The problem says g(x+3) at x=1. Does that mean I find the y value on g where x = 1 (4) and then add 3 and then find g(7)? Or do I add 1 to 3 and take g(4)?
Given [tex]f(x)=3x+2[/tex] and [tex]g(x)=\frac{x-4}{2x}[/tex]
It asks for [tex]g(\frac{1}{x})[/tex]
So substitute in: [tex]\frac{\frac{1}{x} - 4}{2(\frac{1}{x})}[/tex]
Simplify: [tex]\frac{\frac{1}{x} - 4}{\frac{2}{x}}[/tex]
I multiply top and bottom by x right?
That would give: [tex]\frac{1 - 4x}{2}[/tex] and [tex]D_{f} = \Re[/tex] Correct?
------------------
Then on this other problem I have a graph of different functions f and g. The problem says g(x+3) at x=1. Does that mean I find the y value on g where x = 1 (4) and then add 3 and then find g(7)? Or do I add 1 to 3 and take g(4)?