5.5 Increasing and decreasing intervals  (Page 2/2)

Problem : Determine monotonic nature of function :

$$f\left(x\right)=\sqrt{\left(\frac{{\pi }^2}{4}-x^2\right)}$$

Solution : The domain of derivative, f(x), is not R. We first need to find its domain and then determine sign of its derivative within the domain. The expression in the square root in denominator is non-negative. Hence,

$$\Rightarrow \frac{{\pi }^2}{4}-x^2\ge 0$$ $$\Rightarrow x\in [-\frac{\pi }{\mathrm{2,}}\frac{\pi }{2}]$$

Its first derivative is :

$$\Rightarrow f\prime \left(x\right)=-\frac{2x}{2\sqrt{\left(\frac{{\pi }^2}{4}-x^2\right)}}=-\frac{x}{\sqrt{\left(\frac{{\pi }^2}{4}-x^2\right)}}$$

We know that square root is a positive number. It means that sign of derivative will solely depend on the sign of “x” in the numerator. Clearly, derivative is positive for x<0 and negative for x>0. Derivative is zero at x= 0 i.e. at a single point only. But, function is continuous in the given interval. Hence, we include end points as well :

$$\text{Strictly increasing interval}=[-\frac{\pi }{2},0]$$ $$\text{Strictly decreasing interval}=\left[\mathrm{0,}\frac{\pi }{2}\right]$$

Problem : Determine monotonic nature of function :

$$f\left(x\right)=2x^3+3x^2-12x+1$$

Solution : The first derivative of the given polynomial function is :

$$\Rightarrow f\prime \left(x\right)=2X3x^2+3X2x-12=6x^2+6x-12$$

Clearly, the derivative is a quadratic function. We can determine the sign of the quadratic expression, using sign scheme for quadratic expression. Now, the roots of the corresponding quadratic equation when equated to zero is obtained as :

$$\Rightarrow 6x^2+6x-12=0\Rightarrow x^2+x-2=0\Rightarrow x^2+2x-x-2=0$$

$$\Rightarrow x\left(x+2\right)-1\left(x+2\right)=0\Rightarrow \left(x-1\right)\left(x+2\right)=0$$

$$\Rightarrow x=\mathrm{1,}-2$$

Here, coefficient of “ $x^2$ ” is positive. Hence, sign of the middle interval is negative and side intervals are positive.

Since cubic polynomial is a continuous function, we can include end points also in the interval :

$$\text{Strictly increasing interval}=\left(-\infty ,-2\right]\cup \left[\mathrm{1,}\infty \right)$$

$$\text{Strictly decreasing interval}=\left[-\mathrm{2,1}\right]$$

Problem : Determine sub-intervals of [0,π/2] in which given function is (i) strictly increasing and (ii) strictly decreasing.

$$f\left(x\right)=\cos 3x$$

Solution : Its first derivative is :

$$f\prime \left(x\right)=-3\sin 3x$$

Corresponding to given interval [0,π/2], argument to sine function is [0,3π/2]. Sine function is positive in first two quadrants [0, π] and negative in third quadrant [π,3π/2].

Corresponding to these argument values, sin3x is positive in [0,π/3] and negative in [π/3, π/2]of the given interval. But, negative sign precedes 3sin3x. Hence, derivative is negative in [0,π/3] and positive in [π/3, π/2]of the given interval.

Alternatively, we can find zero of sin3x as :

$$-3\sin 3x=0$$ $$\Rightarrow \sin 3x=0$$ $$\Rightarrow 3x=n\pi ;n\in Z$$ $$x=\frac{n\pi }{3};n\in Z$$

Thus, there is one zero at x=π/3 for n=1, in the interval (0,π/2). To test sign, we put x = π/4 in -3sin3x, we have -3sin 3π/4<0. Hence, derivative is negative in [0,π/3] and positive in [π/3, π/2]. Further since sine function is a continuous function, we include end points :

$$\text{Strictly decreasing interval}=\left[\mathrm{0,}\frac{\pi }{3}\right]$$ $$\text{Strictly increasing interval}=\left[\frac{\pi }{\mathrm{3,}}\frac{\pi }{2}\right]$$

Problem : Determine monotonic nature of function :

$$f\left(x\right)=\left(1+x\right)e^x$$

Solution : Its first derivative is :

$$f\prime \left(x\right)=e^x+\left(1+x\right)e^x=\left(x+2\right)e^x$$

Since $e^x$ is positive for all values of x, the derivative is zero if :

$$\left(x+2\right)=0$$ $$\Rightarrow x=-2$$

For a test point, $x=-\mathrm{3,}f\prime \left(x\right)=-1e^{-3}<0$ . Further, function is continuous in R. Hence,

$$\text{Strictly increasing interval}=[-\mathrm{2,}\infty )$$ $$\text{Strictly decreasing interval}=(-\infty ,-2]$$