Ucale

# Increasing and Decreasing Functions

### Introduction

In this chapter, we shall study monotonicity of functions. A function f(x) is said to be monotonicity increasing function on [a, b] if the values of f(x)  increase and decrease with the increase or decrease in x.  If the  values of f(x) decrease with the increase in the values of f(x) then f(x) is said to be monotonicity decreasing function . The monotonicity of a functions in [a, b] is strongly connected to the sign of its derivative in [a, b] . In determining the intervals of monotonicity of a function in its domain, we shall be solving the inequations f(x) > 0 and f (x) <0

### Solution of Rational Algebraic Inequations

The following results are very useful in solving rational algebraic inequations:

(i) ab>0 (a<0 and b>0) or (a<0 and b<0)

(ii) ab<0 (a>0 and b<0) or (a<0 and b<0)

(iii) ab>0 and a>0 = b>0

(iv) ab<0 and a<0 = b>0

### Algorithm to solve rational Algebraic Inequations

If P(x) and Q(x) are polynomials , then the inequations  $\displaystyle \frac { P(x) }{ Q(x) } >0,\frac { P(x) }{ Q(x) } <0,\frac { P(x) }{ Q(x) } \ge 0\quad and\quad \frac { P(x) }{ Q(x) } \le 0$

are known as rational algebraic inequations. These inequations can be solved by using following algorithm.

Step I Factorize P(x) and Q(x) into linear factors.

Step II Make coefficient of x positive in all factors.

Step III Equate all the factors to zero and find the corresponding  values of x. These values are generally known as critical points

Step IV Plot the critical points on the number line. Note that n critical points will divide the number line in (n+1) regions.

Step V In the right most region, the expression will be positive and in other regions it will be alternatively positive and negative. So, mark positive sign n the right most region and than mark alternatively positive and negative sign of remaining regions

Step VI Obtain the solution set of the given inequation by selecting the appropriate regions in Step V

### Example

Solve $\displaystyle 4{ x }^{ 3 }-24{ x }^{ 2 }+44x-24>0$

Solution : We have

$\displaystyle 4{ x }^{ 3 }-24{ x }^{ 2 }+44x-24>0\\ \Rightarrow 4\left( { x }^{ 3 }-6{ x }^{ 2 }+11x-6 \right) \qquad \left[ \because 4>0\quad and\quad ab>0,\quad a>0\Rightarrow b>0 \right] \\ \Rightarrow \left( x-1 \right) \left( { x }^{ 2 }-5x+6 \right) >0\\ \Rightarrow \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) >0\\ \Rightarrow 13\\ \Rightarrow x\quad \epsilon \quad \left( 1,2 \right) \cup \left( 3,\infty \right)$

Hence ,the solution set of the given inequality is $\displaystyle \left( 1,2 \right) \cup \left( 3,\infty \right)$

### Strictly Increasing Function

A function f(x) is said to be a strictly increasing function on (a,b) if

$\displaystyle { x }_{ 1 }<{ x }_{ 2 }\\ \Rightarrow f\left( { x }_{ 1 } \right)

Thus, f(x) is strictly increasing on (a, b) if the values of f(x) increase with the increase in the values of x . Graphically , f(x) is increasing on (a, b) if the graph y=f(x) moves up as x moves to right. The graph in the figure above is graph of a strictly increasing function on (a, b)

### Example

Show that the function f(x)=2 x+3 is strictly increasing function on R

Solution:

$\displaystyle Let\quad { x }_{ 1 },{ x }_{ 2 }\quad \epsilon \quad R\quad and\quad let\quad { x }_{ 1 }<{ x }_{ 2 }\quad Then\\ \Rightarrow { x }_{ 1 }<{ x }_{ 2 }\\ \Rightarrow 2{ x }_{ 1 }<2{ x }_{ 2 }\\ \Rightarrow 2{ x }_{ 1 }+3<2{ x }_{ 2 }+3\\ \Rightarrow f\left( { x }_{ 1 } \right)

So, f(x) is strictly increasing function on R. This result is also evident from the graph of the function shown in the figure below

### Strictly decreasing Function

A function f(x) is said to be a strictly increasing function on (a,b) if

$\displaystyle { x }_{ 1 }<{ x }_{ 2 }\\ \Rightarrow f\left( { x }_{ 1 } \right) >f\left( { x }_{ 2 } \right) \quad \\ for\quad all\quad { x }_{ 1 },{ x }_{ 2 }\quad \epsilon \quad \left( a,b \right)$

Thus, f(x) is strictly decreasing on (a, b) if the values of f(x) decrease with the increase in the values of x . Graphically , it means that  f(x) is decreasing function  on (a, b) if the graph y=f(x) moves up as x moves to right. The graph in the figure above is graph of a strictly decreasing function on (a, b)

### Example

Show that the function f(x) = -3 x+12 is strictly decreasing function on R

Solution: $\displaystyle Let\quad { x }_{ 1 },{ x }_{ 2 }\quad \epsilon \quad R\quad and\quad let\quad { x }_{ 1 }<{ x }_{ 2 }\quad Then\\ \Rightarrow { x }_{ 1 }<{ x }_{ 2 }\\ \Rightarrow -3{ x }_{ 1 }>-3{ x }_{ 2 }\\ \Rightarrow -3{ x }_{ 1 }+12<3{ x }_{ 2 }+12\\ \Rightarrow f\left( { x }_{ 1 } \right) >f\left( { x }_{ 2 } \right) \\ Thus,\quad { x }_{ 1 }<{ x }_{ 2 }\\ \Rightarrow f\left( { x }_{ 1 } \right) >f\left( { x }_{ 2 } \right) \\ for\quad all\quad { x }_{ 1 },{ x }_{ 2 }\quad \epsilon \quad R$

So, f(x) is strictly decreasing function on R

This can also be observed from the graph of the function

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