Q. 2 - Calculus

Question

Solving equations: For each of the following functions g(x), find the solutions of g(x) = 0 and also find the values of x for which g(x) does not exist.

$$\begin{gathered} \left(a\right) g\left(x\right)=\frac{{\displaystyle \frac{1}{2}{x}^{-\frac{1}{2}}\left(1+5x\right)-\sqrt{x}\left(5\right)}}{{\left(1+5x\right)}^2} \\ \left(b\right) g\left(x\right)=2x\sqrt{x-1}+x^2\left(\frac{1}{2}{\left(x-1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right) \\ \left(c\right) g\left(x\right)=\frac{e^x\left(1-e^x\right)-e^x\left(-e^x\right)}{{\left(1-e^x\right)}^2} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The solution of g(x)  when g(x) = 0 is $x=\frac{1}{5}$ and the value of x  for which g(x) does not exist are $x=0 \mathrm{and} x=-\frac{1}{5}.$

Part (b) The solution of g(x)  when g(x) = 0 are $x=0 \mathrm{and} x=\frac{4}{3}$ and the value of x  for which g(x) does not exist is $x=1.$

Part (c) The solution of g(x)  when g(x) = 0 are $x=-7.61 \mathrm{and} x=7.61$ and the value of x  for which g(x) does not exist is $x=0.$

Part (a) Step 1. Given Information.

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

Part (a) Step 2. Solve.

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 x for which g(x) doesn't exist are $0, -\frac{1}{5}.$

Part (b) Step 1. Solve.

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

So,

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

Part (b) Step 2. Solve.

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 x for 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.

Part (c) Step 1. Solve.

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

So,

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

Part (c) Step 2. Solve.

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 at x = 0.