FUNCTIONS

Contents

• Identifying Functions
• Implicit and Explicit functions
• Function Notation
• Algebra of functions
• Composition of functions
• Inverse of Functions
• Summary
• Practice Questions
• Recommended Videos
• Test Questions

Learning Objectives

By the end of this section, you should be able to:

• Describe and identify functions.
• Explain vertical line test.
• Perform different operations on functions.
• Explain and calculate inverse of functions.

Identifying Functions

One of the core concepts in advanced algebra and Precalculus is the function. There are several ways to describe a function, we are going discuss three major description:

Function as a special type of rule/argument:

Mathematically, we have several rules or argument, e.g expressions, formula, functions, etc. 

Equations, like formula, are very broad and can show relation between (two or more) variables such that an input (or independent variable) may be able to produce one or more outputs (or dependent variables); 

but when we talk about function, a function is any kind of rule in which every independent variable (that goes into the rule) must produce only ONE dependent variable (that comes out of the rule).

In other words, function is a special type of rule in which for each input (x), there's only ONE corresponding output (y).

This means that one input cannot produce more than one output in a function.

For instance, the rule y = x² ; is a function because for every value x which enters the rule, there will be only one y. Put x = 4, –4, 7...$\mathrm{\infty }$, each x can only produce one y, nothing more.

But, when we look at y = $\sqrt{\text{x}}$ ;  this is not a function because we would get more than one output y for any value of input x. Let us assume x = 9, values for y (when x is 9) are +3 and –3, this is more than one output value, making us to cancel the relation from being called a function.

Another example of an equation that is not a function is x² + y² = 4 ; we can absolutely get more than one output y for every input x which goes into the function.

In summary: the output of a function must be a single value for each input; an input must not give two or more output.

Quick Review:

Categorise the following equations as function or not a function:

(a) x + y = 1                  

(b) y = x³ – x              

(c) y²= x – 2

(d) x³y = –4

(e) x² – y² = 1               

(f) y = $\frac{\text{x}}{{\text{x}}^2–9}$               

(g) 2xy = 4               

(h) 2x + 3y = 4                   

(i) x² = y²              

(j) x³ + y³ = 4

Solution:

(a) x + y = 1

            y = 1 – x (this is a function because every x will produce only one y).

(b) y = x³ – x (this is a function because every x will not produce more than one y).

(c) y²= x – 2

      y = $\pm \sqrt{\text{x}–2}$ (this is not a function, because what comes out of a square root can either be a positive value and/or a negative value; this means there're 2 output values for any input value ❌).

(d) x³y = –4

         y = $\frac{–4}{{\text{x}}^3}$ (this is a function).

(e) x² – y² = 1

              y² = x² – 1

               y = ±$\sqrt{{\text{x}}^2–1}$ (this is not a function since there can be two y for one x)

(f) y = $\frac{\text{x}}{{\text{x}}^2–9}$ (this is a function).

(g) 2xy = 4

           y = $\frac{2}{\text{x}}$ (this is a function).

(h) 2x + 3y = 4

               3y = 4 – 2x

                 y = $\frac{4–2\text{x}}{3}$ (this is a function).

(i) x² = y²

      y = ± $\sqrt{{\text{x}}^2}$ (this is not a function).

(j) x³ + y³ = 4

             y³ = 4 – x³

               y = $\sqrt[3]{4–{\text{x}}^3}$ (this is a function)

It can be seen from the above review that equations involving even roots (e.g: $\sqrt{\text{input equation}}$, $\sqrt[4]{\text{input equation}}$, etc) are not functions.

Function as a relation in coordinates:

A relation in which each x-coordinate is matched with only one y-coordinate is said to describe y as a function of x.

In other words, a function matches each x-coordinate with only one corresponding y-coordinate.

When a x-coordinate has more than one y-coordinate, the both are not related by a function.

For instance, a set of coordinates: {(–2, 1), (1, 3), (1, 4), (3, –1)} is not related by a function.

Why?

When we take a look at   (–2, 1); x = –2, y = 1

                                              (1, 3); x = 1, y = 3

                                              (1, 4); x = 1, y = 4

                                              (3, –1); x = 3, y = –1

We can see that there are 2 different y-coordinates for x = 1, which annuls a function relation.

However, when we take a look at another set of coordinates: {(–2, 1), (1, 3), (2, 3), (3, –1)}, this is shown to be related by a function, because every x-coordinate has only one y-coordinate.

PRO TIP: in a function, two or more x-coordinates can have the same y-coordinate, but one x-coordinate must not have two or more y-coordinates.

Quick Review:

Categorise the following sets of coordinates into a function or not a function:

(a) {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}                      

(b) {(−3, 0), (1, 6), (2, −3), (4, 2), (−5, 6), (4, −9), (6, 2)}                

(c) {(−3, 0), (−7, 6), (5, 5), (6, 4), (4, 9), (3, 0)}                

(d) {(1, 2), (4, 4), (9, 6), (16, 8), (25, 10), (36, 12), . . .}

Solution:

(a) {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} is a function because no x-coordinate occurred twice having different y-coordinate.

(b) {(−3, 0), (1, 6), (2, −3), (4, 2), (−5, 6), (4, −9), (6, 2)} is not a function; x-coordinate "4" occurred twice with different y-coordinates.

(c) {(−3, 0), (−7, 6), (5, 5), (6, 4), (4, 9), (3, 0)} is a function.

(d) {(1, 2), (4, 4), (9, 6), (16, 8), (25, 10), (36, 12), . . .} is a function.

Vertical Line Test

Before a graphical relation can be valid as a function, it must pass the vertical line test. 

A vertical line test is carried out by drawing vertical lines all along (every section on) a curve (or you can picture a vertical line moving to and fro along a curve). 

If a vertical line crosses (any) two points, simultaneously, on a curve, then the graphical relation is not a function, 

but if no vertical line is able to cross any two points on a curve, then the graphical relation is a function.

Determining whether a graph is of a function or not using vertical line test 

Function as mapping of sets

The standard description of function is with respect to sets. 

When we consider two sets, A and B, A contains elements while B also contains elements; a function is then said to be any rule that relates all the elements in set A, to elements in set B such that ONE element in set A is related to only ONE element in set B. This means that each element in set A is related to its single, corresponding element in set B.

One element in set A will not relate to two elements in set B, if this happens, then the both sets are not related by a function.

It is noteworthy that in functions, it is not impossible for different elements in set A to relate to a single element in set B.

The elements of set A are often called inputs, and the elements of the set B are called outputs. 

A function takes an input and converts it into a single output.

Functions are also called maps, mappings, or transformation, while the mapping of elements in one set to elements in another set is shown in a mapping diagram.

The relation of one element a, in set A, to an element b in set B is depicted as: f(a) = b which reads as b is a function of a. We shall see more of this in function notation.

Implicit and Explicit functions

Consider two functions y = 2x² – 3 and 2x² – y – 3 = 0; 

y serves as the output variable, and is a function of input x: [f(x)]. The former function is an explicit function, we normally say it is because the output (y) is written in terms of the input (x); the latter function is known as an implicit function as it is written in terms of both the input and output .

Quick Review: 

Categorise these equations into implicit or explicit functions:

(a) x + y = 1                  

(b) y = x³ – x              

(c) y = $\sqrt{\text{x}–2}$       

(d) x³y = –4

(e) x² – y² = 1               

(f) y = $\frac{\text{x}}{{\text{x}}^2–9}$              

(g) 2xy = 4

(h) 2x + 3y = 4                   

(i) x² = y²              

(j) x³ + y³ = 4

Solution

(a) x + y = 1 is an implicit function.

(b) y = x³ – x is an explicit function.

(c) y = $\sqrt{\text{x}–2}$ is an explicit function.

(d) x³y = –4 is an implicit function.

(e) x² – y² = 1 is not a function.

(f) y = $\frac{\text{x}}{{\text{x}}^2–9}$ is an explicit function.

(g) 2xy = 4 is an implicit function.

(h) 2x + 3y = 4 is an implicit function.

(i) x² = y² is not a function.

(j) x³ + y³ = 4 is an implicit function.

Function Notation

It is widely known that mathematicians don't write too many words, symbols are used instead.

We talked about a function, that it relates the elements of one set to the elements of another set (mathematically, we may deal with possibly the same set).

There is a notation, that is used to represent this situation:

if the function name is f , and the input name is x ,then the unique corresponding output is called f(x) (which is read as " f of x ".)

Thus, f(x) = output of any rule involving input x

We can also use others like g(x), h(x) or simply y.

Without writing much, let us see examples :)

Example 1

What does the function notation g(7) represent?

Solution

g(7) depicts the output from the function g when the input is 7.

Example 2

Suppose f(x) = x + 2 . What is f(3) ?

Solution

f(x) = x + 2

f(3) = ?

We simply substitute the occurrence of 'x' with '3'

f(3) = 3 + 2

f(3) = 5

Example 3

Suppose f(x) = x + 2 . What is f(x + 5) ?

Solution

f(x) = x + 2

f(x + 5) = we replace x with x + 5

f(x + 5) = (x + 5) + 2

f(x + 5) = x + 7

Example 4

If f(x) = –x² + 3x + 4

Find and simplify the following:

(a) f(–1),  f(0),  f(2)

(b) f(2x),  2f(x)

(c) f(x + 2),  f(x) + 2,  f(x) + f(2)

Solution

(a) f(–1):

To find f(–1) we replace every occurrence of x in the expression f(x) with –1

      f(–1) = – (–1)² + 3(–1) + 4

      f(–1) = –(1) –3 + 4

      f(–1) = –1 –3 + 4

      f(–1) = 0

Similarly;

       f(0):

      f(0) = –(0)² + 3(0) + 4

      f(0) = –(0) + 0 + 4

      f(0) = –0 + 0 + 4

      f(0) = 4

       f(2):

 

      f(2) = –(2)² + 3(2) + 4

      f(2) = –4 + 6 + 4

      f(2) = 6

(b) f(2x):

To find f(2x) we replace every occurrence of x in the expression f(x) with 2x

     f(2x) = –(2x)² + 3(2x) + 4

     f(2x) = –4x² + 6x + 4

     2f(x):

2f(x) means that we should multiply f(x) by 2

     2f(x) = 2 × (–x² + 3x + 4)

     2f(x) = –2x² + 6x + 8

(c) f(x + 2);

      f(x + 2) = –(x + 2)² + 3(x + 2) + 4

      f(x + 2) = –(x² + 4x + 4) + 3(x + 2) + 4

      f(x + 2) = –x² –4x –4 + 3x + 6 + 4

      f(x + 2) = –x² – x + 6

     f(x) + 2;

For f(x) + 2, we simply add 2 to the expression f(x)

     f(x) + 2 = [–x² + 3x + 4] + 2

     f(x) + 2 = –x² + 3x + 6

 

   f(x) + f(2);

   f(x) = –x² + 3x + 4

   f(2) = –(2)² + 3(2) + 4 = 6

   f(x) + f(2) = [–x² + 3x + 4] + [6]

   f(x) + f(2) = –x² + 3x + 10

Example 5

If r(x) = $\frac{2\text{x}}{{\text{x}}^2–9}$

Calculate for r(–1), r(0), r(2), r(3)

Solution

r(–1) = $\frac{2(–1)}{(–1{)}^2–9}$

r(–1) = $\frac{–2}{1–9}$

r(–1) = $\frac{–2}{–8}$

r(–1) = $\frac{1}{4}$

r(0) = $\frac{2(0)}{(0{)}^2–9}$

r(0) = $\frac{0}{0–9}$

r(0) = 0

r(2) = $\frac{2(2)}{(2{)}^2–9}$

r(2) = $\frac{4}{4–9}$

r(2) = $\frac{4}{–5}$

r(2) = –$\frac{4}{5}$

r(3) = $\frac{2(3)}{(3{)}^2–9}$

r(3) = $\frac{6}{9–9}$

r(3) = $\frac{6}{0}$

Anything divided by 0 is undefined (in a calculator, you will get a math error), hence, x = 3 is not one of the inputs that can go into the function $\frac{2\text{x}}{{\text{x}}^2–9}$. 

Before now, you may have thought that any value for x could go into any function to produce an output, but that is not the case because we are usually careful of the inputs that go into a function, which is what brings us to Domain, Codomain, and Range– to be discussed in the next blog post.

Algebra of Functions

Like terms, two or more functions can be combined by addition, subtraction, multiplication or division.

Example 6

Given f(x) = 2x + 1  and g(x) = x² + 2x – 1, find (f + g)(x) and (f + g)(2).

Solution

(f + g)(x) is another way of writing f(x) + g(x), it implies that we add the functions f(x) and g(x) together:

(f + g)(x) = f(x) + g(x)

(f + g)(x) = [2x + 1] + [x² + 2x – 1]

(f + g)(x) = x² + 4x

Similarly;

(f + g)(2) = f(2) + g(2)

We can either use what we got when we simplified (f + g)(x):

(f + g)(x) = x² + 4x

(f + g)(2) = 2² + 4(2)

(f + g)(2) = 4 + 8 = 12

Or we can calculate step by step:

(f + g)(x) = [2x + 1] + [x² + 2x – 1]

(f + g)(2) = [2(2) + 1] + [2² + 2(2) – 1]

(f + g)(2) = [4 + 1] + [4 + 4 – 1]

(f + g)(2) = [5] + [7]

(f + g)(2) = 12

Example 7

Given f(x) = 2x – 5  and g(x) = 1 – x, find (f – g)(x) and (f – g)(2)

Solution

(f – g)(x) = f(x) – g(x)

(f – g)(x) = [2x – 5] – [1 – x]

(f – g)(x) = 2x – 5 – 1 + x

(f – g)(x) = 2x + x – 5 – 1

(f – g)(x) = 3x – 6

Therefore;

(f – g)(2) = 3(2) – 6

(f – g)(2) = 6 – 6

(f – g)(2) = 0

Example 8

Given f(x) = x² + 1 and g(x) = x – 4 , find (fg)(x) and (fg)(3).

Solution

(fg)(x) implies that we multiply f(x) by g(x)

(fg)(x) = f(x) × g(x)

(fg)(x) = [x² + 1] × [x – 4]

(fg)(x) = x² [x – 4] + 1[x – 4]

(fg)(x) = x³ – 4x² + x – 4

Therefore;

(fg)(3) = 3³ – 4(3)² + 3 – 4

(fg)(3) = 27 – 4(9) – 1

(fg)(3) = 27 – 36 – 1

(fg)(3) = –10

Example 9

Given f(x) = x + 1  and  g(x) = x – 1, find  $\left(\frac{f}{g}\right)(\text{x})$ and $\left(\frac{f}{g}\right)(3)$

Solution

$\left(\frac{f}{g}\right)(\text{x})$ = $\frac{f(\text{x})}{g(\text{x})}$

$\left(\frac{f}{g}\right)(\text{x})$ = $\frac{\text{x}+1}{\text{x}–1}$

All real numbers ($\mathbb{R}$) can enter this function as an input except 1. 1 would render this function as invalid (because we will get 0 as denominator, which leads to undefined), we therefore add x $\ne$ 1; x must not be equal to 1 for this function!

We can calculate for $\left(\frac{f}{g}\right)(\text{3})$ as:

$\left(\frac{f}{g}\right)(\text{3})$ = $\frac{3+1}{3–1}$

$\left(\frac{f}{g}\right)(\text{3})$ = $\frac{4}{2}$

$\left(\frac{f}{g}\right)(\text{3})$ = 2

Composition of Functions

Consider two functions f(x) and g(x), we can combine these functions by writing one in terms of the other.

In other words, we can combine two functions by setting one function as an input inside another function. 

For instance, if f(x) = x + 1 and g(x) = x – 1, we can write f(x) in terms of g(x) such that the function g(x) would replace all occurrence of the input value x in f(x);

This becomes f(g(x)) = (x – 1) + 1; 

g(x) has replaced the input x in f(x).

This is known as composition of functions.

It is the creation of a new function by putting one function inside another function.

 

Making use of f(x) and g(x), writing f(x) in terms of g(x) is shown as f(g(x)), read as f of g of x. 

It is also written as ($\text{f}\circ \text{g}$)(x).

When this happens, the function g(x) would replace every occurrence of input x in f(x).

Let us see examples :)

Example 10

Given f(x) = x² and g(x) = x + 1 , find ($\text{f}\circ \text{g}$)(x) and ($\text{g}\circ \text{f}$)(x).

Solution

($\text{f}\circ \text{g}$)(x):

Since ($\text{f}\circ \text{g}$)(x) = f(g(x)), then;

($\text{f}\circ \text{g}$)(x) = f(x + 1)

                = (x + 1)²

($\text{g}\circ \text{f}$)(x):

Since ($\text{g}\circ \text{f}$)(x) = g(f(x)), then;

($\text{g}\circ \text{f}$)(x) = g(x²)

                = (x²) + 1

                = x² + 1

Note that ($\text{f}\circ \text{g}$)(x) $\ne$ ($\text{g}\circ \text{f}$)(x).

Example 11

Find ($\text{f}\circ \text{g}$)(3) and ($\text{g}\circ \text{f}$)(3) if f(x) = x + 2 and g(x) = 4 – x²

Solution

($\text{f}\circ \text{g}$)(3):

Since ($\text{f}\circ \text{g}$)(x) = f(g(x))

Then, ($\text{f}\circ \text{g}$)(x) = f(4 – x²)

                          = (4 – x²) + 2

                          = 6 – x²

Evaluating for ($\text{f}\circ \text{g}$)(3);

($\text{f}\circ \text{g}$)(3) = 6 – 3²

                = 6 – 9

                = –3

($\text{f}\circ \text{g}$)(3) = –3

($\text{g}\circ \text{f}$)(3):

Since ($\text{g}\circ \text{f}$)(x) = g(f(x))

Then, ($\text{g}\circ \text{f}$)(x) = g(x + 2)

                           = 4 – (x + 2)²

                          = 4 – (x² + 4x + 4)

                          = 4 – x² – 4x – 4

                          = –x² – 4x

Evaluating for ($\text{g}\circ \text{f}$)(3);

($\text{g}\circ \text{f}$)(3) = –(3)² – 4(3)

                = –9 – 12

               = –21

Inverse of Functions

While functions are rules that map an input to an output, inverse of functions does the opposite. Inverse of a function takes an output and maps it to its corresponding input. 

In other words, inverse of a function aims to find the original input (of a particular function) for which only its output is known. Conditions have to be met before we can say that a function is invertible- we will see that in the next lesson.

Inverse of a function is denoted as f ^–1, and it is used to calculate for the unknown input of a known output.

Consider a function f(x) = x + 1, how do we get f ^–1(x)?

There's a quick method for doing this; 

we transform the output of the function, f(x), to an input, x, while the input of the function, x, which is about to be calculated, becomes f ^–1(x)– then we make f ^–1(x) the subject of the equation.

This becomes:

x = f ^–1(x) + 1

f ^–1(x) = x – 1

It is important to note that many inverse of functions may not be functions themselves. The inverse of a function that can produce only one value for every input is known as inverse function.

Example 12

Find the inverse of the function f(x) = 3x – 2

Solution

Step 1: transform f(x) to x; and x to f ^–1(x).

x = 3(f ^–1(x)) – 2

3(f ^–1(x)) = x + 2

f ^–1(x) = $\frac{\text{x}+2}{3}$

Example 13

Find an unknown input which produced an output of 8 from the function f(x) = $\frac{6\text{x}}{\text{x}–1}$

Solution

Solving for the inverse of this function;

Step 1: transform f(x) to x; and x to f ^–1(x)

x = $\frac{6f^{–1}\text{(x)}}{f^{–1}\text{(x)}–1}$

x × $\left(f^{–1}\text{(x)}–1\right)$ = 6 $f^{–1}(\text{x})$

xf ^–1(x) – x = 6f ^–1(x)

6f ^–1(x) – xf ^–1(x) = –x

f ^–1(x) (6 – x) = –x

f ^–1(x) = $\frac{–\text{x}}{6–\text{x}}$

f ^–1(x) = $\frac{\text{x}}{\text{x}–6}$

Thus; the input which gave an output 8 is calculated as;

f ^–1(x) = $\frac{8}{8–6}$

f ^–1(x) = $\frac{8}{2}$

f ^–1(x) = 4

Therefore, it is calculated that an input variable '4' produced an output variable '8' via the function f(x) = $\frac{6\text{x}}{\text{x}–1}$

Summary

📌 A function is any rule that produces only one output for each input.

📌 A vertical line test is used to see if a curve is the curve of a function or not, any graph that a vertical line can cross at any two points, is not the graph of a function.

📌 Functions are also called maps, mappings, or transformation.

📌 Implicit functions are written in terms of both input and output, while explicit functions are written in terms of input only.

📌 A function relation is shown as f(a) = b, which reads as b is a function of a.

Practice Questions

Question 1

Given that $f(x)=3x²+2x–5$, determine $f(0)$

A. $5$

B. $0$

C. $–5$

D. $1$

Question 2

Determine $f(2)$ in $f(x)$ = $\frac{x^2-x}{x-2}$ and state your conclusion.

A. Undefined, the function cannot have an input value of $2$

B. $0$, the function cannot have an input value of $2$

C. $1$, the function cannot have an input value of $2$

D. $1$, the function can have an input value of $2$

Question 3

Given that $f(x)=5x^2+x–7$; determine $\frac{f(2)}{f(1)}$

A. $15$

B. $–15$

C. $7.5$

D. $–7.5$

Question 4

If $f(x)=x^2-x$ and $\text{g}(x)=x+1$, determine $\text{f}\circ \text{g}(2)$

A. $4$

B. $6$

C. $9$

D. $3$

Question 5

Given that $f(x)=3x^2+2x$ and $\text{g}(x)=x^2+1$, determine $(\frac{\text{g}\circ \text{f}(2)}{\text{f}\circ \text{g}(2)}+\frac{\text{f}\circ \text{g}(2)}{\text{g}\circ \text{f}(2)})\times \text{g}\circ \text{f}\circ \text{g}(2)$

A. $7226$

B. $24237.95$

C. $19706.72$

D. $55.47$

Question 6

Find $f^{-1}(x)$ if $f(x)=x^2-x$ and state if $f^{-1}(x)$ is a function or not.

A. $f^{-1}(x)=\sqrt{\frac{x}{x+1}}$ This is a function

B. $f^{-1}(x)=\sqrt{\frac{x}{x+1}}$ This is not a function

C. $f^{-1}(x)=\frac{1\pm \sqrt{1+4x}}{2}$ This is a function

D. $f^{-1}(x)=\frac{1\pm \sqrt{1+4x}}{2}$ This is not a function

Question 7

Find $\frac{f(x+h)-f(x)}{h}$ given that $f(x)=x^2-x+1$

A. $2x-1+h$

B. $\frac{2x-h+1}{h}$

C. $\frac{x^2}{h}$

D. $1$

Question 8

Determine $\text{f}\circ \text{g}\circ \text{h}(5)$ if $f(x)=2x+1$     $\text{g}(x)=x^2$     and     $h(x)=x+3$

A. $64$

B. $129$

C. $124$

D. $120$

Question 9

Given that $f(x)=\frac{1}{2}(3^x+3^{-x})$   and   $\text{g}(x)=\frac{1}{2}(3^x-3^{-x})$. Determine $f(0),\text{g}(0)$ and $f^{-1}(x)$

A. $1,0,\frac{x\pm \sqrt{x^2-1}}{3}$

B. $0,1,\frac{\log (x\pm \sqrt{x^2-1})}{\log 3}$

C. $1,0,\frac{\log (x\pm \sqrt{x^2-1})}{\log 3}$

D. $0,1,\frac{x\pm \sqrt{x^2-1}}{3}$

Question 10

Given that $f(x)=\frac{1}{2}(3^x+3^{-x})$   and   $\text{g}(x)=\frac{1}{2}(3^x-3^{-x})$. Simplify $f(u+v)f(u)-\text{g}(u+v)\text{g}(u)$ and comment on your answer.

A. $\frac{1}{2}(3^u+3^{-u})$ ; it can be seen that $f(u+v)f(u)-\text{g}(u+v)\text{g}(u)=f(u)$

B. $\frac{1}{2}(3^v+3^{-v})$ ; it can be seen that $f(u+v)f(u)-\text{g}(u+v)\text{g}(u)=f(v)$

C. $\frac{1}{2}(3^u-3^{-u})$ ; it can be seen that $f(u+v)f(u)-\text{g}(u+v)\text{g}(u)=g(u)$

D. $\frac{1}{2}(3^v-3^{-v})$ ; it can be seen that $f(u+v)f(u)-\text{g}(u+v)\text{g}(u)=g(v)$

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FUNCTIONS

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FUNCTIONS

Total Questions: 20

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Quiz Answers

1. Function f is defined by f(x) = –2x² + 6x – 3 find f(–2)

–23

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2. Functions f and g are defined by f(x) = –7x – 5 and g(x) = 10x – 12 find (f + g)(2)

–11

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3. Given that f(x) = x² – 2x + 1 and g(x) = (x – 1)(x + 3) find (^f/_g)(x)

^{(x – 1)}/_{(x + 3)}

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4. Evaluate f(3) given that f(x) = |x – 6| + x² – 1

11

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5. Given that h(x) = x² – 4x + 9, calculate for x when h(x) = 30

–3 or 7

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6. Find (f o g)(4) given that f(x) = √(x) and g(x) = x² – 2x + 1

3

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7. Find an unknown input which gave an output of 12 from the function f(x) = ^6x/_{x – 1}

2

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8. If f(x) = 3x + 2 and g(x) = x² – 1, find f(g(–3))

26

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9. Given f(x) = x³ + 1 and g(x) = 2x – 5, calculate for h(–2) if h(x) = [f(x)]² – g(x)

58

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10.
Which of the above diagrams represent functions?

b, e

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11. Given that f(x) = ^x/_{x + 3} and g(x) = ²/_x, calculate (f o g)(²/₃)

¹/₂

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12. Given that f(x) = ln(4x – 2), find an expression for f ^–1(x)

¹/₄ (e^x + 2)

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13. If f(x) = x² + 1, find an expression for f ^–1(x)

f ^–1(x) = √(x – 1)

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14. Given that f(x) = 4 – ln(2x – 1), determine f(f(1))

4 – ln7

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15. Given that f(x) = √(x + 4), find an expression for f ^–1(x)

f ^–1(x) = x² – 4

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16. If f(x) = ¹/₂e^x + 1, find an expression for f ^–1(x)

f ^–1(x) = ln(2x – 2)

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17. Given that f(x) = (x – 3)² + 1 for which x ≥ 4. Calculate x when f(x) = 17

7

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18. Given that f(x) = 3x – 2, solve the equation f(x) = f ^–1(x)

1

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19. Given that f(x) = e^x and g(x) = x² + 1, find g(f(x))

e^2x + 1

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20. Given that f(x) = x² + 3 and g(x) = 2x + 2, calculate (f o g)(x) = 2(g o f)(x) + 15

3

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