91Ó°ÊÓ

The given function is$g\left(x\right)=\frac{{\displaystyle \frac{1}{2}{x}^{-\frac{1}{2}}\left(1+5x\right)-\sqrt{x}\left(5\right)}}{{\left(1+5x\right)}^2}.$

To find the solutions of g(x) = 0,let's simplify it.

So,

$\begin{array}{rl}g\left(x\right)& =\frac{\frac{(1+5x)}{2\sqrt{x}}−\sqrt{x}\left(5\right)}{(1+5x{)}^2}\\ g\left(x\right)=& \frac{\frac{(1+5x)−2\left(5\right)\sqrt{x}\sqrt{x}}{2\sqrt{x}}}{(1+5x{)}^2}\\ g\left(x\right)=& \frac{1+5x−10x}{2\sqrt{x}(1+5x{)}^2}\\ g\left(x\right)=& \frac{1−5x}{2\sqrt{x}(1+5x{)}^2}\end{array}$

Now, let's find g(x)=0,

$$\begin{gathered} g\left(x\right)=0 \\ \begin{array}{c}\frac{1−5x}{2\sqrt{x}(1+5x{)}^2}=0\\ 1−5x=0\\ x=\frac{1}{5}\end{array} \end{gathered}$$

Now, the value of xfor whichg(x)doesn't exist are$0,-\frac{1}{5}.$

To find the solutions of g(x) = 0,let's simplify it.

So,

$$\begin{gathered} g\left(x\right)=2x\sqrt{x-1}+x^2\left(\frac{1}{2}{\left(x-1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right) \\ g\left(x\right)=2x\sqrt{x−1}−\frac{x^2}{2\sqrt{x−1}} \\ g\left(x\right)=\frac{2x\sqrt{x−1}\left(2\sqrt{x−1}\right)−x^2}{2\sqrt{x−1}} \\ \begin{array}{c}g\left(x\right)=\frac{4x(x−1)−x^2}{2\sqrt{x−1}}\\ g\left(x\right)=\frac{4x^2−4x−x^2}{2\sqrt{x−1}}\\ g\left(x\right)=\frac{3x^2−4x}{2\sqrt{x−1}}\\ g\left(x\right)=\frac{x(3x−4)}{2\sqrt{x−1}}\end{array} \end{gathered}$$

Now, let's find g(x)=0,

$$\begin{gathered} g\left(x\right)=0 \\ \begin{array}{rl}\frac{x(3x−4)}{2\sqrt{x−1}}& =0\\ x(3x−4)& =0\\ x=0\mathrm{and}& x=\frac{4}{3}\end{array} \end{gathered}$$

Now, to find the value of xfor which g(x)doesn't exist put a denominator equal to zero,

$$\begin{gathered} 2\sqrt{x-1}=0 \\ x=1 \end{gathered}$$

Thus the function doesn't exist at x = 1.

To find the solutions of g(x) = 0,let's simplify it.

So,

$\begin{array}{rl}g\left(x\right)& =\frac{e^x(1−e^x)−e^x(−e^x)}{{(1−e^x)}^2}\\ g\left(x\right)& =\frac{e^x−e^{2x}+e^{2x}}{{(1−e^x)}^2}\\ g\left(x\right)& =\frac{e^x}{{(1−e^x)}^2}\\ g\left(x\right)& =e^x{(1−e^x)}^{−2}\end{array}$

Now, to find g(x)=0,we will draw the graph

So, the g(x)=0,when $x=-7.61\mathrm{and}x=7.61.$

From the graph, we can depict that the function is doesn't exist atx = 0.