## GRE Functions

Our chapter on GRE functions enables you to learn each and every minuscule detail about GRE functions and covers problems ranging from easy to difficult.

An algebraic expression of only one variable may be defined as a function, usually f or g, of that variable.

Example:   What is the value of the function $$f(x)\quad =\quad { x }^{ 2 }-1$$

When x = 1?

Solution:   In the function $$f(x)\quad =\quad { x }^{ 2 }-1$$, if x = 1, then f(1)=1-1=0. In other words, when the input to the function is 1, the output is 0.

The set of input numbers for a function is the domain & the set of output values is the range of function. Every input value of a function has exactly one output value.

However, more than one input value may have the same output value.

Consider the function $$f(x)\quad =\quad { x }^{ 2 }-1$$ again.

If $$x\quad =\quad -1,f(-1)\quad =\quad { (-1) }^{ 2 }-1\quad =\quad 0$$

If $$x\quad =\quad 1,\quad f(1)\quad =\quad { (1) }^{ 2 }-1\quad =\quad 0$$

Therefore,$$f(1)=f(-1)=0$$ .

Restricted domain of a function:

A function may be defined for all real numbers or it may be only for a subset of the real number.

For example: $$g(x)\quad =\quad -{ x }^{ 2 }+2$$ , where $$-3\le x\le 3$$

The domain of the function $$g(x)$$ is restricted to values of $$x$$ between -3 and +3 inclusive.

The maximum value of $$g(x)\quad =\quad -{ x }^{ 2 }+2$$ occurs when $$x=0,\quad g(0)=\quad -{ x }^{ 2 }+2\quad =\quad 0+2\quad =\quad 2.$$

To find the minimum value of $$g(x)\quad =\quad -{ x }^{ 2 }+2$$, where $$-3\le x\le 3$$

We will find g(x) at extreme values.

$$g(3)\quad =\quad -9+2\quad =\quad -7,\quad g(3)\quad =\quad -9+2\quad =\quad -7$$

Therefore, minimum value occurs when x = $$-3$$ or x = 3. That value is -7.

The domain of a function may also be restricted to avoid having a zero in the denominator of a fraction or to avoid taking the square root of a negative number.

Example: $$f(x)\quad =\quad \frac { 2x-5 }{ x-3 } \quad$$, where $$x\quad \neq \quad 3$$

Here, To find the domain of the function, we are supposed to restrict domain of f to avoid x-3=0.

Example:  $$h(x)\quad =\quad \sqrt { x+10 } ,\quad where\quad x\ge -10$$

Here, To find the domain of this function, we are supposed to restrict domain of h to avoid x+10<0.

## Graphs of functions:

To graph, a function in the xy-plane, use the x-axis for the input & the y-axis for the output. You can represent every input value, x & its corresponding output value y, as an ordered pair (x,y).

Problem 1) 1) Plot function $$f(x)\quad =\quad 2x-1\quad and\quad g(x)\quad =\quad { x }^{ 2 }-1$$

2) Find points of intersection of two graphs.

Sol.   We can easily see that three graph intersect at two points. These are the points for which f(x)=g(x).

$$2x\quad -1\quad =\quad { x }^{ 2 }-1\\ { x }^{ 2 }-2x\quad =\quad 0\\ x(x\quad -\quad 2)\quad =\quad 0\\ x=0\quad or\quad x=2.\\ f(0)\quad =\quad -1,\quad f(2)\quad =\quad 3\\$$

Therefore, the point of intersection is (0,1) & (2,3).

## Piecewise function:

A piecewise function is defined by more than one equation, where each equation applies to a different part of the domain of the function.

The absolute value function is an example of a piecewise of a piecewise function.

The absolute value function is defined as

$$f(x)\quad =\quad |x|\quad =\\ \{ \quad x\quad for\quad x\ge 0\\ \{ \quad x\quad for\quad x<0$$

## Shifting of graphs: –

The ability to visualize how graphs shift when the basic analytical expression is changed is a very important skill.

In order to be able to do so, you first need to understand the following points clearly.

• The graph of f(x)+c is the graph of f(x) shifted upward c units or spaces.
• The graph of f(x)-c is the graph of f(x) shifted downward c units or spaces.
• The graph of f(x+c) is the graph of f(x) shifted to the left c units or spaces.
• The graph of f(x-c) is the graph of f(x) shifted to the right c units or spaces.

Problem 2) The graph of $$f(x)\quad =\quad { x }^{ 2 }$$ is shown. How is the graph of the function $$g(x)\quad =\quad { x }^{ 2 }-4$$  related to the graph of f(x)?

Sol.

Graph of $$f(x)\quad =\quad { x }^{ 2 }$$

Graph of $$g(x)\quad =\quad { x }^{ 2 }-4$$

As we have studied that the graph of f(x)-c is obtained by shifting graph of f(x) c units downward.

Problem 3) The graph of $$h(x)\quad =\quad |x|$$ is shown. How is the graph of the function $$k(x)\quad =\quad |x-1|$$ related to the graph of $$h(x)$$ ?

Sol.

As we have studied that the graph of f(x-c) is the graph of f(x) shifted to the right c units or spaces.

The graph of k(x)=|x-1| is obtained by shifting graph of h(x)=|x| 1 unit to the right.

The graph of a function may also be vertically stretched away or compressed toward the x-axis by a factor of c, where c is a positive number. You can stretch the graph by making the slope larger & steeper, or you can compress the graph by making the slope smaller & less steep. To change the vertical slope of the graph of f(x), follow these rules:

• The graph of c×f(x) is the graph of f(x) stretched away from the x-axis by a factor of c.
• The graph of (1/c)×f(x) is the graph of f(x) compressed away from the x-axis by a factor of c.

Problem 4) describe the relationship between the graphs of $$f(x)\quad =\quad { x }^{ 2 }$$  & $$g(x)\quad =\quad \frac { 1 }{ 2 } { x }^{ 2 }$$.

Sol.  As we have studied that the graph of  is the graph of (1/c)×f(x) is the graph of f(x) compressed toward the x-axis by a factor of c.

The graph of $$g(x)\quad =\quad \frac { 1 }{ 2 } { x }^{ 2 }$$ is obtained by compressing the graph of f(x) toward the x-axis by a factor of 2.

## PROBLEMS

Problem 5) For what values should be the domain be restricted for this function?

$$f(n)\quad =\quad \frac { n-6 }{ { n }^{ 2 }-6n }$$

Sol.  Factor the denominator: $${ n }^{ 2 }-6n\quad =\quad n(n-6)$$

As a denominator of 0 is not permissible, the value of n cannot be 6 or 0.

Problem 6) If $$g(a)\quad =\quad { a }^{ 2 }-1$$ & f(a)=a+4, what is g(f(-4))?

Sol.

$$f(a)\quad =\quad a+4\\ f(-4)\quad =\quad a\quad +\quad 4\\ -4\quad +\quad 4\quad =\quad 0.\\ g(0)\quad =\quad 0-1\quad =\quad -1$$

Problem 7) If $$f(x)\quad =\quad \frac { 2x+6 }{ 4 }$$ and [/latex]g(x)\quad =\quad 2x-1[/latex], what is f(g(x))?

Sol.

$$f(x)\quad =\quad \frac { 2x+6 }{ 4 } \\ g(x)\quad =\quad 2x-1\\ f(g(x))\quad =\quad \frac { 2(2x-1)+6 }{ 4 } \\ (Replace\quad x\quad with\quad 2x-1)\\ =\quad (\frac { 4x-2+6 }{ 4 } )\\ =\quad \frac { 4x+4 }{ 4 }$$

Problem 8) If $$g(x)\quad =\quad \frac { { x }^{ 2 }(4x+9) }{ (3x-3)(x+2) }$$, for which of the following x values is g(x) undefined?

Indicate all such values of x.

a) $$-\frac { 9 }{ 4 }$$

b) -2

c) 0

d) 1

e) 2

Sol.  Factor the denominator:$$\quad \quad (3x-3)(x+2)\\ =\quad 3(x-1)(x+2)$$

As a denominator of 0 is not permissible, the value of x cannot be 1 or -2.

The correct answers are -2 & 1.

Problem 9) If $$f(x)\quad =\quad x+5$$ & $$f(2g)\quad =\quad 19$$. What is the value of f(3-g)?

Sol.

$$f(x)\quad =\quad x+5\\ f(2g)\quad =\quad 2g+5\quad =\quad 19\\ g\quad =\quad 7\\ f(3-g)\quad =\quad f(3-7)\quad =\quad f(-4)\quad =\quad -4+5\quad =\quad { 1 }^{ Ans }$$

Problem 10) If $$f(x)\quad =\quad { x }^{ 2 }-1$$, what is the value of f(y)+f(-1)?

Sol.

$$f(x)\quad =\quad { x }^{ 2 }-1\\ f(y)\quad =\quad { y }^{ 2 }-1\\ f(-1)\quad =\quad { (-1) }^{ 2 }-1\quad =\quad 0\\ f(y)\quad +\quad f(-1)\quad =\quad { y }^{ 2 }-1+0\\ =\quad { ({ y }^{ 2 }-1) }^{ Ans }$$

Problem 11) If $$\ast x$$ is defined as the square of one-half of x, what is the value of $$\frac { \ast 5 }{ \ast 3 }$$ ?

Sol.  $$\ast x$$ is defined as the square of one-half of x, $$\ast x\quad =\quad { (\frac { x }{ 2 } ) }^{ 2 }$$

Therefore,

$$\ast 5\quad =\quad { (\frac { 5 }{ 2 } ) }^{ 2 }=\quad \frac { 25 }{ 4 } \\ \ast 3\quad =\quad { (\frac { 3 }{ 2 } ) }^{ 2 }=\quad \frac { 9 }{ 4 } \\ \frac { \ast 5 }{ \ast 3 } \quad =\quad (\frac { \frac { 25 }{ 4 } }{ \frac { 9 }{ 4 } } )\quad =\quad \frac { 25 }{ 9 }$$

Problem 12) If $$h(x)\quad =\quad 5{ x }^{ 2 }+x$$, then which of the following is equal to h(a+b)?

a) $$5{ a }^{ 2 }+5{ b }^{ 2 }$$

b) $$5{ a }^{ 3 }+5{ b }^{ 3 }$$

c) $$5{ a }^{ 2 }+5{ b }^{ 2 }+a+b$$

d) $$5{ a }^{ 3 }+10ab+5{ b }^{ 3 }$$

e) $$5{ a }^{ 2 }+10ab+5{ b }^{ 2 }+a+b$$

Sol.

$$h(x)\quad =\quad 5{ x }^{ 2 }+x\\ h(a+b)\quad =\quad 5{ (a+b) }^{ 2 }+(a+b)\\ =\quad 5{ a }^{ 2 }+5{ b }^{ 2 }+5(2ab)+a+b\\ =\quad 5{ a }^{ 2 }+10ab+5{ b }^{ 2 }+a+b$$

Problem 13) $$\ast x$$ is defined as the least integer greater than x for all odd values of x, & the greatest integer less than x for all even values of x. What is the value of $$\ast (-2)-\ast 5$$?

Sol. $$\ast x$$ is defined as the least integer greater than x for all odd values of x, & the greatest integer less than x for all even values of x.

If x is odd, $$\ast x$$ equals the least integer greater than x (e.g., if x=3, then the “least integer greater than 3” is equal to 4).

If x is even, $$\ast x$$ equals the greatest integer less than x (e.g., if x=6, the “greatest integer less than x” is equal to 5).

Since -2 is even, $$\ast (-2)$$ = the greatest integer less than -2, or -3.

Since 5 is odd, $$\ast (5)$$ = the least integer greater than 5, or 6.

$$\ast (-2)-\ast (5)\quad =\quad -3-6\quad =\quad { (-9) }^{ Ans }$$