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2 2. For function h(x) - a. Find (-1) b. Find the domain of h(x), write your answer in set-builder notation. 2 c. For functio
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Answer #1

2.

Solution:

a.

\small h\left ( x \right )=\frac{2}{x+3}

Replace \small x by  \small -1

\small \therefore h\left ( -1 \right )=\frac{2}{-1+3}=\frac{2}{2}=1

..h(-1)=1

b.

The domain is the set of input or argument values for which the function is real and defined.

Find undefined points:  

\small x+3=0\Rightarrow x=-3

The function domain is :

\small \left \{ x\in \mathbb{R}\: |\: x<-3\: \: \mathrm{or}\: \: x>-3 \right \}

\small c.

To find the inverse function,swap \small x and \small y , and solve the resulting equation for \small x.

\small y=\frac{2}{x+3} becomes \small x=\frac{2}{y+3}

Now, solve the equation  \small x=\frac{2}{y+3} for \small y .

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

\small \therefore h^{-1}\left ( x \right )=-3+\frac{2}{x}

Singular point: \small x=0

Domain :  \small \left ( -\infty ,0 \right )\cup \left ( 0,\infty \right )

The function range is the combined domain of the inverse functions.

Inverse of the function :

\small \frac{2}{x+3}

Domain of the inverse function is:

\small x<-3 or \small x>-3

Combine the intervals:

\small f\left ( x \right )<-3 or   \small f\left ( x \right )>-3

i.e.  \small \left ( -\infty ,-3 \right )\cup \left ( -3,\infty \right )

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