Monotonicity, Maxima and Minima
4.0 Finding intervals of increasing and decreasing functions
4.0 Finding intervals of increasing and decreasing functions
1. After differentiation the function, if we get $f'(x)$ in the form of $\frac{{P(x)}}{{Q(x)}}$ in which both $P(x)$ and $Q(x)$ are algebraic functions. In this case we use number line method to find the intervals in which the function is increasing or decreasing.
2. After differentiation the function, if we get $f'(x)$ a trigonometric function. We use the graphical method to find the intervals in which the function is increasing or decreasing.
3. Interval of increasing and decreasing functions should lie with in the domain of the given function.
Question 6. Find the monotonicity of the function $$f(x) = x\ln x$$
Solution: To find the intervals in which the function is increasing or decreasing, we should first differentiate it with respect to $x$, we get $$f'(x) = 1 + \ln x$$ To find the interval in which it is monotonically increasing, we can write $$\begin{equation} \begin{aligned} f'(x) \geqslant 0 \\ 1 + \ln x \geqslant 0 \\ \ln x \geqslant - 1 \\ x \geqslant {e^{ - 1}} \\ x \geqslant \frac{1}{e} \\\end{aligned} \end{equation} $$ To find the interval in which it is monotonically decreasing, we can write $$\begin{equation} \begin{aligned} f'(x) \leqslant 0 \\ 1 + \ln x \leqslant 0 \\ \ln x \leqslant - 1 \\ x \leqslant {e^{ - 1}} \\ x \leqslant \frac{1}{e} \\\end{aligned} \end{equation} $$ Therefore, the function is monotonically increasing in the interval $x \in \left[ {\frac{1}{e},\infty } \right)$ and is monotonically decreasing in the interval $x \in \left( {\infty ,\frac{1}{e}} \right]$.
Question 7. Find the monotonicity of the function $$f(x) = 2{x^3} - 9{x^2} + 12x + 15$$
Solution: To find the intervals in which the function is increasing or decreasing, we should first differentiate it with respect 
to $x$, we get $$\begin{equation} \begin{aligned} f'(x) = 6{x^2} - 18x + 12 \\ f'(x) = 6({x^2} - 3x + 2) \\ f'(x) = 6({x^2} - 2x - x + 2) \\ f'(x) = 6\left\{ {x(x - 2) - 1(x - 2)} \right\} \\ f'(x) = 6(x - 2)(x - 1) \\\end{aligned} \end{equation} $$ To find the intervals we use the number line method as shown in figure. From this, we can say that $f'(x)>0$ or increasing in the interval $\left( { - \infty ,1} \right] \cup \left[ {2,\infty } \right)$ and $f'(x)<0$ or decreasing in the interval $[1,2]$.
to $x$, we get $$\begin{equation} \begin{aligned} f'(x) = 6{x^2} - 18x + 12 \\ f'(x) = 6({x^2} - 3x + 2) \\ f'(x) = 6({x^2} - 2x - x + 2) \\ f'(x) = 6\left\{ {x(x - 2) - 1(x - 2)} \right\} \\ f'(x) = 6(x - 2)(x - 1) \\\end{aligned} \end{equation} $$ To find the intervals we use the number line method as shown in figure. From this, we can say that $f'(x)>0$ or increasing in the interval $\left( { - \infty ,1} \right] \cup \left[ {2,\infty } \right)$ and $f'(x)<0$ or decreasing in the interval $[1,2]$.Question 8. Find the monotonicity of the function $$f(x) = \frac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}}\quad ;\;x \in [0,2\pi ]$$
Solution: To find the intervals in which the function is increasing or decreasing, we should first differentiate it with respect 
to $x$, we get $$\begin{equation} \begin{aligned} f'(x) = \frac{{(2 + \cos x)(4\cos x - 2 + x\sin x - \cos x) + \sin x(4\sin x - 2x - x\cos x)}}{{{{\left( {2 + \cos x} \right)}^2}}} \\ f'(x) = \frac{{4\cos x - 4 + 3 + {{\sin }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}} \\ f'(x) = \frac{{4\cos x - 1 + 1 - {{\cos }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}} \\ f'(x) = \frac{{\cos x(4 - \cos x)}}{{{{\left( {2 + \cos x} \right)}^2}}} \\\end{aligned} \end{equation} $$ In the denominator, term will always positive and in the numerator, $4-cosx$ will always be positive as the value of $cosx$ when $x \in \left[ {0,2\pi } \right]$ is always between $\left[ { - 1,1} \right]$. Therefore, the sign of $f'(x)$ will depend only on the sign of $cosx$. We apply the graphical method to find the intervals as shown in figure.
to $x$, we get $$\begin{equation} \begin{aligned} f'(x) = \frac{{(2 + \cos x)(4\cos x - 2 + x\sin x - \cos x) + \sin x(4\sin x - 2x - x\cos x)}}{{{{\left( {2 + \cos x} \right)}^2}}} \\ f'(x) = \frac{{4\cos x - 4 + 3 + {{\sin }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}} \\ f'(x) = \frac{{4\cos x - 1 + 1 - {{\cos }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}} \\ f'(x) = \frac{{\cos x(4 - \cos x)}}{{{{\left( {2 + \cos x} \right)}^2}}} \\\end{aligned} \end{equation} $$ In the denominator, term will always positive and in the numerator, $4-cosx$ will always be positive as the value of $cosx$ when $x \in \left[ {0,2\pi } \right]$ is always between $\left[ { - 1,1} \right]$. Therefore, the sign of $f'(x)$ will depend only on the sign of $cosx$. We apply the graphical method to find the intervals as shown in figure. The function is increasing in the interval $\left[ {0,\frac{\pi }{2}} \right] \cup \left[ {\frac{{3\pi }}{2},2\pi } \right]$ and it is decreasing in the interval $\left[ {\frac{\pi }{2},\frac{{3\pi }}{2}} \right]$.
Note: We have considered the interval of $[0,2\pi]$ because the interval of increasing and decreasing fucntions should lie with in the domain of the given function and in the question there is a condition on $x$.
