Given:
[tex]g(x)=\frac{x+\sqrt{x}}{x+1}[/tex]
Find-: Domain, intersection with axis, limit at end the domain, draw the graph, first derivative the minimum and maximum of the point is:
Sol:
Domain:
The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined. In the function machine metaphor, the domain is the set of objects that the machine will accept as inputs.
[tex]\begin{gathered} g(x)=\frac{x+\sqrt{x}}{x+1} \\ \\ Domain:x>0 \end{gathered}[/tex]
The domain is greater than the zero "0" because the negative value of "x" is create undefine form inside the square root.
Intersection with axis:
For x-intersection value of "y" is zero so,
[tex]\begin{gathered} g(x)=\frac{x+\sqrt{x}}{x+1} \\ \\ 0=\frac{x+\sqrt{x}}{x+1} \\ \\ 0=x+\sqrt{x} \\ \\ \sqrt{x}(\sqrt{x}+1)=0 \\ \\ x=0,\sqrt{x}=-1 \\ \\ \sqrt{x}=-1\text{ Not possible } \\ x=0 \end{gathered}[/tex]
For y intersection value of "x" is zero then,
[tex]\begin{gathered} y=\frac{x+\sqrt{x}}{x+1} \\ \\ y=\frac{0+\sqrt{0}}{0+1} \\ \\ y=0 \end{gathered}[/tex]
Graph of function is:
The first derivative is:
[tex]\begin{gathered} g(x)=\frac{x+\sqrt{x}}{x+1} \\ \\ g^{\prime}(x)=\frac{(x+1)(1+\frac{1}{2\sqrt{x}})-(x+\sqrt{x)}}{(x+1)^2} \\ \\ g^{\prime}(x)=\frac{x+\frac{\sqrt{x}}{2}+1+\frac{1}{2\sqrt{x}}-x-\sqrt{x}}{(x+1)^2} \\ \\ g^{\prime}(x)=\frac{1+\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}}{(x+1)^2} \\ \\ \end{gathered}[/tex]
For maximum is:
[tex]\begin{gathered} g^{\prime}(x)=0 \\ \\ 1+\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}=0 \\ \\ \frac{2\sqrt{x}+1-x}{2\sqrt{x}}=0 \\ \\ x-1=2\sqrt{x} \end{gathered}[/tex]
Square both side then:
[tex]\begin{gathered} (x-1)^2=2\sqrt{x} \\ \\ x^2+1-2x=4x \\ \\ x^2-6x+1=0 \\ \\ x=\frac{6\pm\sqrt{36-4}}{2} \\ \\ x=\frac{6\pm\sqrt{32}}{2} \\ \\ x=5.83,x=0.17 \end{gathered}[/tex]
For the maximum or minimum point are
(5.83) and 0.17
