5.7 Least and greatest function values  (Page 3/3)

Problem 5: A function is defined as :

$$f\left(x\right)=\frac{e^x}{1+\left[x\right]}$$

If domain of the function is [0,∞), find the range of the function.

Solution :

Statement of the problem : The function contains greatest integer function in its denominator, whereas numerator of functon is an exponential function. Greatest integer function, GIF, is defined in discontinuous intervals of 1. It returns integral value equal to the starting value of discontinuous interval. The function, however, is continuous in sub-intervals of 1 i.e. in a finite interval. It is clear that function values and function nature will change in accordance with the values of GIF in sub-intervals of the domain. We need to determine nature of function in few of the initial sub-intervals and extrapolate the results to determine the range of function as required.

In order to determine nature of function, we write function and its derivative in the initial sub-intervals of the given domain. Note that domain of function starts with 0.

$$0\le x<\mathrm{1,}f\left(x\right)=e^x,f\prime \left(x\right)=e^x$$

$$1\le x<\mathrm{2,}f\left(x\right)=\frac{e^x}{2},f\prime \left(x\right)=\frac{e^x}{2}$$

$$2\le x<\mathrm{3,}f\left(x\right)=\frac{e^x}{3},f\prime \left(x\right)=\frac{e^x}{3}$$

$$\text{-------------------------------------}$$

The derivatives in each interval are positive for values of “x” in the domain. It means that function is strictly increasing in each of the intervals. There is no minimum and maximum as monotonic nature of function does not change. The least and greatest values, therefore, correspond to end function values in each finite sub-intervals. Using these least and greatest values, we determine range of initial sub-intervals as given here :

$$0\le x<1\Rightarrow e^0\le y<e^1\Rightarrow 1\le y<e$$

$$1\le x<2\Rightarrow \frac{e}{2}\le y<\frac{e^2}{2}$$

$$2\le x<3\Rightarrow \frac{e^2}{3}\le y<\frac{e^3}{3}$$

$$\text{-------------------------------------}$$

The plot of the function is drawn in the figure. Note that function values in adjacent intervals overlap each other, but exceeds the greatest value in the preceding intervals. Clearly, function value begins from “1” and continues towards infinity. Hence, range of the function is :

$$\left[\mathrm{1,}\infty \right)$$

Problem : Find range of continuous function :

$$f\left(x\right)=\sqrt{\left(x-1\right)}+\sqrt{\left(5-x\right)}$$

Solution : Given function is continuous. Also, it involves square root of polynomial. It means that function is defined in sub interval of real number set R. We, therefore, need to find the domain of the function first. Then we investigate the nature of the function in its domain and determine least (a) and greatest (b) function values.

For individual square roots to be real,

$$x-1\ge 0\Rightarrow x\ge =1$$

$$5-x\ge 0\Rightarrow x\le 5$$

Hence, domain of the function is intersection of individual domains and is a finite interval [1,5]. Now, we differentiate the function with respect to independent variable to determine the nature of function :

$$\Rightarrow f\prime \left(x\right)=\frac{1}{2\sqrt{\left(x-1\right)}}-\frac{1}{2\sqrt{\left(5-x\right)}}$$

In order to draw sign scheme, we find the roots of the equation by putting f’(x) = 0 and solving for “x” as :

$$\Rightarrow 2\sqrt{\left(x-1\right)}=2\sqrt{\left(5-x\right)}$$

$$\Rightarrow x-1=5-x\Rightarrow 2x=6\Rightarrow x=3$$

For test value, x = 4, f’(x) is negative. Hence, corresponding sign scheme of f’(x) is as given here :

Since function is increasing to the left and decreasing to the right of root value, the function value at root value is maximum. Clearly, there is only one maximum. Thus, it is greatest value also. Further, either of two function values at end points is least value of the function in the domain . Now, function values at end points and root point are :

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

$$\Rightarrow f\left(3\right)=\sqrt{2}+\sqrt{2}=2\sqrt{2}$$

$$\Rightarrow f\left(5\right)=2+0=2$$

Clearly, function at x = 3 is greatest value and either values at x = 1 and 5 is least. Hence, range of the function is :

$$\left[\mathrm{2,}2\sqrt{2}\right]$$