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# Suppose f(x)=\frac{3}{2-x}. If g(x)=\frac{1}{f^{-1}(x)}+9, find g(3).

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Suppose f(x)=\frac{3}{2-x}. If g(x)=\frac{1}{f^{-1}(x)}+9, find g(3).

Nov 7, 2019

#1
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Bro post in rendered latex please! Stop being lazy buddy!

Suppose $$f(x)=\frac{3}{2-x}$$. If $$g(x)=\frac{1}{f^{-1}(x)}+9$$, find $$g(3)$$.

Find the inverse of function f(x).     You do this by setting up f(x) as y, then switch "x" and "y". Solve for y.

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$$y=\frac{3}{2-x}$$

$$x=\frac{3}{2-y}$$

$$2x-yx=3$$

$$2-y=\frac{3}{x}$$

$$y=-\frac{3}{x}+2$$

$$y=2-\frac{3}{x}$$

So now we figured out f-1(x). We then plug that into g(x).

$$g(x)=\frac{1}{2-\frac{3}{x}}$$

We then evaluate g(3).

I will leave that up to you now

Nov 7, 2019
edited by CalculatorUser  Nov 7, 2019

#1
+3

Bro post in rendered latex please! Stop being lazy buddy!

Suppose $$f(x)=\frac{3}{2-x}$$. If $$g(x)=\frac{1}{f^{-1}(x)}+9$$, find $$g(3)$$.

Find the inverse of function f(x).     You do this by setting up f(x) as y, then switch "x" and "y". Solve for y.

.

$$y=\frac{3}{2-x}$$

$$x=\frac{3}{2-y}$$

$$2x-yx=3$$

$$2-y=\frac{3}{x}$$

$$y=-\frac{3}{x}+2$$

$$y=2-\frac{3}{x}$$

So now we figured out f-1(x). We then plug that into g(x).

$$g(x)=\frac{1}{2-\frac{3}{x}}$$

We then evaluate g(3).

I will leave that up to you now

CalculatorUser Nov 7, 2019
edited by CalculatorUser  Nov 7, 2019