Inequalities of Combined Functions

Examples and Answers

Example 1Let and

a) Graph the functions on the same set of axes. Identify the points of intersection.

b) Illustrate the region for which:

i) ii)

xxf )( 2)2()( xxg

)()( xgxf )()( xfxg

SolutionThe graph of is a line with slope equal to one, passing through the origin. The graph of

can be obtained by applying a horizontal translation of 2 units to the right of the graph

)(xfy

2xy

2)2()( xxg

From the graph, the points of intersection appear to be (1,1) and (4,4). This can be verified algebraically by solving the linear-quadratic system of f(x) and g(x). Set the two functions equal and solve for x.)()( xgxf

2)2( xx

442 xxx

0452 xx

0)1()1( xx

01x1x

04 x4x

To verify that the points of intersection are (1,1) and (4,4), mentally substitute each of these ordered pairs into and

Method 1: Compare the Functions VisuallyUsing different colours or line styles is a useful technique for visualizing regions for which one function is greater than another. The graph of y= f(x) is above the graph of y= g(x) between x =1 and x =4. Therefore,

i) f(x) >g(x) on the interval (1,4)ii) g(x) >f(x) on the interval (- ∞,1) U (4,∞)

xxf )( 2)2()( xxg

Method 2: Analyze the Difference FunctionAnother way to illustrate where one function is greater then another is to subtract the functions and see where the graph of difference is above the x-axis.

The function, ,is positive on the intervals where and therefore

)()( xgxf

0)()( xgxf2)2()()( xxxgxf

)44( 2 xxx

452 xx

452 xxy

0)()( xgxf )()( xgxf

This function is positive on the interval (1,4). Therefore,

i) on the interval (1,4)ii) on the intervals (-,1) U (4,)

Method 3: Analyze the Quotient Function

The function f(x) and g(x) can also be compared by analyzing their quotient. Graph the combined function and identify the intervals for which this quotient is greater than one, which will correspond to where

)()( xgxf )()( xfxg

2)2()(

)(

x

x

xg

xf

)()( xgxf

To make this more clearly visible, graph the combined function using a graphing calculator with an appropriate viewing window. Include a graph of y=1. Use the Intersect operation to identify the coordinates of the points of intersection.

It appears that on the interval (1,4). However,

is not defined for .Determine and to decide what happens when

Thus, andi) on the interval (1,4)ii) on the intervals (-,1) U (4, )

1)(

)(

xg

xf

)(

)(

xg

xf2x )2(f

)2(g 2x

2)2( f 0)2( g

)()( xfxg

)()( xgxf

Example 2Solve

This is a problem that cannot be solved by using algebra; it must be solved using a graphing calculator in a few simple steps:

xx logsin

Solution• graph and • make sure the calculator is in Radian mode• change the window settings• use the Intersection operation to find the x-coordinates of any points of intersection– there is an asymptote at – there are points of intersection at (2.70, 0.43), (7.33,

0.87), (8.26, 0.92)   

xy sin xy log

0x

- the graph of is above the graph of - between and , and between and

- therefore: (0, 2.70) U (7.33, 8.26)- this can be verified by using the TEST in the

calculator menu- enter the inequality in the Y=- set calculator to Dot mode- where the equation is true, a line will appear on

the graph

xy sin xy log0x 70.2x 33.7x

26.8x

xx logsin

Example 3A computer store’s cost, C, for shipping and storing n computers can be modeled by the function

The storage capacity of the store’s warehouse is 750 units.

a) Graph this function and explain its shape. What is the domain of interest for this problem?

b) Determine the maximum and minimum number of computers that can be ordered at any time to keep costs below $1,500, assuming that inventory has fallen below zero.

c) What is the optimum order size that will minimize storage costs?

d) Why might this not be the best number to order?

nnnC

2000005.1)(

Solutiona) Use a graphing calculator. Apply number sense

to choose appropriate window settings.

b) The shape of this graph can be understood if it is thought of as the superposition of a linear function and a rational function.

The domain of is {nER, n>0}b) Determine the values of n that will keep costs

below $1500.

)(nC

nnnC

2000005.1)(

Method 1: Graphing CalculatorGraph the adjusted cost function

and identify the region below the horizontal axis. Apply number sense to choose the region below the horizontal axis. Apply number sense to choose appropriate window settings to view the region of interest.

1500200000

5.1 n

nC

Notice that the adjustment term translates the original cost curve down 1,500 units, making it easy to distinguish the region of interest. Locate the zeros by using the Zero operation.

According to the graph, an inventory order of form 169 to 841 units will keep the costs below $1,500. This region can be illustrated by setting Y2=0 and then shading above and below for the line styles of Y1 and Y2, respectfully.

The region where the two shaded parts overlap shows the interval for which costs are below $1,500.

Method 2: Use Pencil and Paper

The quadratic expression on the left side corresponds to a quadratic function that opens upward. The region between the zeros will correspond to n-values for which . Solve for these zeros.

015002000005.1 2 nn

020000015005.1 2 nn

020000015005.1 2 nn

a

acbbn

2

42

)5.1(2

)200000)(5.1(4)1500()1500( 2 n

3

1050000)1500 n

842158 or

0C

Since the warehouse has a maximum storage capacity of 750 units, a shipment of from 159 to 750 computers will keep costs below $1,500.

c) To determine the optimum order size that will minimize shipping and storage costs, locate the minimum of the original cost function,

Use the Minimum operation of the graphing calculator.

nnnC

2000005.1)(

The minimum cost occurs when approximately 356 computers are ordered. The corresponding cost for this size of order is approximately $1,095.

d) The optimum order size may not be desirable, particularly if consumer demand is not aligned with this value. If demand significantly exceeds 365 units, then profit will be lost due to unfilled sales requests. On the other hand, if demand is significantly below 365 units, unnecessary shipping and store costs will occur, as well as the cost of inventory space for other products.