Monotonicity, Maxima and Minima
5.0 Proving inequality using monotonicity
5.0 Proving inequality using monotonicity
We have to prove the given inequality using the concepts explained above.
First of all, using the inequality we have to make a function $f(x)$, then we differentiate it and check whether for the given intervals of $x$, it is positive or negative in the given interval.
Then find the value of the function at $x=0$ and $x = \infty $ and use the values according to the monotonicity of the function.
If it is not possible to understand whether function is increasing or decreasing after differentiating it first time i.e., from $f'(x)$. Then, differentiate it again and check whether $f''(x)$ is positive or negative in the given intervals and use it to solve the inequality and so on as explained in the questions.
Question 9. Using monotonicity, prove that $$\tan x > x + \frac{{{x^3}}}{3}\quad ;\;x \in \left( {0,\frac{\pi }{2}} \right)$$
Solution: First step is to make a function from the inequality. We can write $$f(x) = \tan x - x - \frac{{{x^3}}}{3}$$ Differentiating it, we get $$f'(x) = {\sec ^2}x - 1 - {x^2}$$ But we can not say whether $f'(x)$ is positive or negative in the given interval. So, differentiating it once again, we get $$f''(x) = 2{\sec ^2}x\tan x - 2x$$ Again from this, we can not say whether $f''(x)$ is positive or negative in the given interval. So, differentiating it once again, we get $$\begin{equation} \begin{aligned} f'''(x) = 4{\sec ^2}x{\tan ^2}x + 2{\sec ^4}x - 2 \\ f'''(x) = 4{\sec ^2}x{\tan ^2}x + 2{\left( {{{\sec }^2}x} \right)^2} - 2 \\ f'''(x) = 4{\sec ^2}x{\tan ^2}x + 2{\left( {1 + {{\tan }^2}x} \right)^2} - 2 \\ f'''(x) = 4{\sec ^2}x{\tan ^2}x + 2\left( {1 + {{\tan }^4}x + 2{{\tan }^2}x} \right) - 2 \\ f'''(x) = 4{\sec ^2}x{\tan ^2}x + 2 + 2{\tan ^4}x + 4{\tan ^2}x - 2 \\ f'''(x) = 4{\sec ^2}x{\tan ^2}x + 2{\tan ^4}x + 4{\tan ^2}x \\\end{aligned} \end{equation} $$ Now, it is clear that $f'''(x)$ is positive for all values of $x \in \left( {0,\frac{\pi }{2}} \right)$. Since $f'''(x)>0$, we can say that $f''(x)$ is an increasing function and also $$f''(0) < f''(x)\;;\;x \in \left( {0,\frac{\pi }{2}} \right)$$ and $$\begin{equation} \begin{aligned} f''(0) = 2{\sec ^2}0\tan 0 - 2.0 \\ f''(0) = 0 \\\end{aligned} \end{equation} $$ Therefore, we can write $$f''(x) > 0$$ This means $f'(x)$ is also an increasing function. Using the same concept again, we can write $$f'(0) < f'(x)\;;\;x \in \left( {0,\frac{\pi }{2}} \right)$$ and $$f'(0) = {\sec ^2}0 - 1 - {0^2} = 1 - 1 = 0$$ Therefore, we can write $$f'(x)>0$$ This means $f(x)$ is also an increasing function. Using the same concept again, we can write $$f(0) < f(x)\;;\;x \in \left( {0,\frac{\pi }{2}} \right)$$ and $$f(0) = \tan 0 - 0 - \frac{{{0^3}}}{3} = 0$$ Therefore, we can write $$\begin{equation} \begin{aligned} f(x) > 0 \\ \tan x - x - \frac{{{x^3}}}{3} > 0 \\ \tan x > x + \frac{{{x^3}}}{3} \\\end{aligned} \end{equation} $$
Question 10. Find which one of the following is maximum. $${(1)^{\frac{1}{1}}},{(2)^{\frac{1}{2}}},{(3)^{\frac{1}{3}}},{(4)^{\frac{1}{4}}},{(5)^{\frac{1}{5}}},{(6)^{\frac{1}{6}}},{(7)^{\frac{1}{7}}}$$
Solution: Let us assume the function be $$f(x) = {(x)^{\frac{1}{x}}}$$ Let us find the intervals in which the function is 
increasing and decreasing. Differentiating it with respect to $x$, we get $$f(x) = y = {x^{\frac{1}{x}}}$$ Taking log both side, we get $$\begin{equation} \begin{aligned} \log y = \log {\left( x \right)^{\frac{1}{x}}} \\ \log y = \frac{1}{x}\log x \\\end{aligned} \end{equation} $$ Differentiating w.r.t. $x$, we get $$\begin{equation} \begin{aligned} \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{x}.\frac{1}{x} - \frac{{\log x}}{{{x^2}}} \\ \frac{{dy}}{{dx}} = y\left( {\frac{{1 - \log x}}{{{x^2}}}} \right) = {x^{\frac{1}{x}}}\left( {\frac{{1 - \log x}}{{{x^2}}}} \right) \\ \frac{{dy}}{{dx}} = {x^{\frac{1}{x} - 2}}(1 - \log x) \\ \frac{{dy}}{{dx}} = f'(x) = {x^{\frac{{1 - 2x}}{x}}}(1 - \log x) \\\end{aligned} \end{equation} $$ Now, it is increasing function when $$\begin{equation} \begin{aligned} f'(x) \geqslant 0 \\ {x^{\frac{{1 - 2x}}{x}}}(1 - \log x) \geqslant 0 \\ 1 - \log x \geqslant 0 \\ \log x \leqslant 1 \\ 0 < x \leqslant e \\\end{aligned} \end{equation} $$ and it is decreasing function when $$\begin{equation} \begin{aligned} f'(x) \leqslant 0 \\ {x^{\frac{{1 - 2x}}{x}}}(1 - \log x) \leqslant 0 \\ 1 - \log x \leqslant 0 \\ \log x \geqslant 1 \\ x \geqslant e \\\end{aligned} \end{equation} $$ Therefore, the function is increasing in the interval $(0,e]$ and decreasing in the interval $[e,\infty )$ as shown in figure. From the graph, we can say that the maximum value will be either at ${(2)^{\frac{1}{2}}}$ or ${(3)^{\frac{1}{3}}}$. We can write $${(2)^{\frac{1}{2}}} = {(2)^{\frac{1}{3} \times \frac{3}{2}}} = {\left( {{{(8)}^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} = {\left( {2\sqrt 2 } \right)^{\frac{1}{3}}} = {(2.82)^{^{\frac{1}{3}}}}$$ On comparing, we can say that the maximum value is of ${(3)^{\frac{1}{3}}}$.
increasing and decreasing. Differentiating it with respect to $x$, we get $$f(x) = y = {x^{\frac{1}{x}}}$$ Taking log both side, we get $$\begin{equation} \begin{aligned} \log y = \log {\left( x \right)^{\frac{1}{x}}} \\ \log y = \frac{1}{x}\log x \\\end{aligned} \end{equation} $$ Differentiating w.r.t. $x$, we get $$\begin{equation} \begin{aligned} \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{x}.\frac{1}{x} - \frac{{\log x}}{{{x^2}}} \\ \frac{{dy}}{{dx}} = y\left( {\frac{{1 - \log x}}{{{x^2}}}} \right) = {x^{\frac{1}{x}}}\left( {\frac{{1 - \log x}}{{{x^2}}}} \right) \\ \frac{{dy}}{{dx}} = {x^{\frac{1}{x} - 2}}(1 - \log x) \\ \frac{{dy}}{{dx}} = f'(x) = {x^{\frac{{1 - 2x}}{x}}}(1 - \log x) \\\end{aligned} \end{equation} $$ Now, it is increasing function when $$\begin{equation} \begin{aligned} f'(x) \geqslant 0 \\ {x^{\frac{{1 - 2x}}{x}}}(1 - \log x) \geqslant 0 \\ 1 - \log x \geqslant 0 \\ \log x \leqslant 1 \\ 0 < x \leqslant e \\\end{aligned} \end{equation} $$ and it is decreasing function when $$\begin{equation} \begin{aligned} f'(x) \leqslant 0 \\ {x^{\frac{{1 - 2x}}{x}}}(1 - \log x) \leqslant 0 \\ 1 - \log x \leqslant 0 \\ \log x \geqslant 1 \\ x \geqslant e \\\end{aligned} \end{equation} $$ Therefore, the function is increasing in the interval $(0,e]$ and decreasing in the interval $[e,\infty )$ as shown in figure. From the graph, we can say that the maximum value will be either at ${(2)^{\frac{1}{2}}}$ or ${(3)^{\frac{1}{3}}}$. We can write $${(2)^{\frac{1}{2}}} = {(2)^{\frac{1}{3} \times \frac{3}{2}}} = {\left( {{{(8)}^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} = {\left( {2\sqrt 2 } \right)^{\frac{1}{3}}} = {(2.82)^{^{\frac{1}{3}}}}$$ On comparing, we can say that the maximum value is of ${(3)^{\frac{1}{3}}}$.