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### Introduction

Function: A relationship between a set of inputs and a set of allowable outputs is defined as a function and each input is is related to a particular output. A function is also described as a relation between different variables. It is commonly expressed as an expression involving one or more variables.
The inputs to a function are variables such as $x$.
The output of the function can be represented as $f(x)$ or $g(x)$ for a particular value of $x$.
Output of the function is always equal to two times the value of $x$ ie.
$f(x)$ = $x$
Eg: If $x$ = 1 then $f(x)$ = 2$x$.
Quadratic function: $f(x)$ = $ax^2 + bx + c$
Where, $a$, $b$, and $c$ are constants and $a$ ≠ 0
This form of function $f(x)$ is known as a quadratic function. On the coordinate plane, if a quadratic equation is drawn then the result is a parabola. This can be either open upward or downward U-shaped curve.

### Methods

Estimation of functions:
Function estimation means calculating $f(x)$ at some particular value $x$.
Eg: $f(x)$ = $x^2$ - 1. Then find $f(2)$?
Substitute $x$ = 2, then
$f(2)$ = $2^2$ - 1
$f(2)$ = 3
Operations on functions:
Functions can be added, subtracted, multiplied and divided like any other quantity. Some of the below listed key rules will make these operations simpler.
Rules:
For any two functions of $f(x)$ and $g(x)$
Addition: $(f + g)(x)$ = $f(x)$ + $g(x)$
Subtraction: $(f - g)(x)$ = $f(x)$ - $g(x)$
Multiplication: $(f * g)(x)$ = $f(x)$ * $g(x)$
Division: $\frac{f}{g}(x)$ = $\frac{f(x)}{g(x)}$; $g(x)$ ≠ 0
Compound function:
A function that performs on another function is known as a compound function.
It is represented as "$f(g(x))$".
First evaluate internal function $g(x)$ then outer function $f(x)$. It's double substitution.
Domain and range:
Domain of functions: The set of inputs($x$ values) for which the function is defined is defined as a domain of a function.
How to find domain of a function:
First look for any restrictions on the domain, to calculate the domain of the function. There are two main restrictions for function domain. They are:
• Division by zero: It is impossible mathematically. Eg: $f(x)$ = $\frac{1}{x - 2}$ is undefined at $x$ = 2, since when $x$ = 2, the function $f(x)$ = $\frac{1}{0}$. Therefore, a function is therefore undefined for all the values of $x$ for which division by zero occurs.

• Negative numbers under square roots: The square roots of a negative number does not exist, so if a function contains a square root, such $f(x)$ = $\sqrt{x}$, the domain must be $x$ > 0.

Range of a function: The set of all values of $f(x)$ that can be generated by the function is known as a range.

• The easiest way to find range is to visualize it on a graph.

• $x$-axis consists of all the values of $x$ of domain.

• $y$-axis consists of all the values of $f(x)$ of range.

There are two main signs of functions with limited ranges. They are absolute value and even exponents.
Absolute value: The absolute value of a quantity is always positive.
Eg: If $f(x)$ = $\mid{x}\mid$
$f(x)$ must be always positive.
So, the range includes only zero and positive numbers i.e. range is $f(x)$ ≥ 0.
Never expect that any function with an absolute value symbol has the same range.
The range of $g(x)$ = $\mid{x}\mid$ is zero and all the negative quantity will be positive i.e. $f(x)$ ≤ 0.
Even exponents: By squaring a number (or raise it to any multiple of 2) at any time the resulting quantity will be positive.
How to find range of a function:
Figure A

Figure B

• It is similar to finding of domain.

• First, look for absolute values, even exponents or other reasons that the range would be restricted.

• Adjust the range step by step to find the domain.

• The graphs of equations of the form $y$ = $ax^2$ are examples of parabola.

• These parabolas are symmetric about the $y$ axis; the open upward and have a lowest point at (0, 0) if $a$ > 0 in (figure A) and the open downward have a highest point at (0, 0) if a < 0 (figure B).

• The highest or lowest point of the graph of $y$ = $ax^2$ is called the vertex of the parabola and its line of symmetry is known as the axis of symmetry as simply the axis of the parabola.

### Samples

1. Find the vertex and axis of symmetry of the graph of $f(x)$ = -3$x^2$ - 12$x$ - 1 and sketch the graph.
Solution:

2. Find the vertex and the $y$ and $x$ intercepts of the graph of $f(x)$ = $-2x^2 - 5x + 3$. Determine whether the graph open upward or downward and sketch the graph?
Solution:
Given
$f(x)$ = $-2x^2 - 5x + 3$
Here, $a$ = -2, $b$ = -5, and $c$ = 3
By vertex formula $h$ = $\frac{-b}{2a}$
= $\frac{-(-5)}{2(-2)}$
= $\frac{-5}{4}$
$k$ = $f(h)$ = $f(\frac{-5}{4})$
= -2$(\frac{-5}{4})^2$ - 5$\frac{-5}{4}$ + 3
= $\frac{49}{8}$
The $y$ intercept is $c$ = 3
The $x$ intercept are solutions of equation i.e.
-2$x^2$ - 5$x$ + 3 = 0
= ($x$ + 3)(-2$x$ + 1) = 0
$x$ = -3 and $\frac{1}{2}$
As $a$ = -2 < 0, the graph opens downward as shown below.

3. If $\frac{3x}{4 - x}$, find the value of $f(x + 1)$?
Solution:
Given that,
$f(x)$ = $\frac{3x}{4 - x}$
Now, substitute $(x + 1)$ in the place of $x$ i.e.
$f(x + 1)$ = $\frac{3(x + 1)}{4 - (x + 1)}$
$f(x + 1)$ = $\frac{3x + 3}{4 - x - 1}$
$f(x + 1)$ = $\frac{3x + 3}{3 - x}$
Therefore, the value of $f(x + 1)$ = $\frac{3x + 3}{3 - x}$

4. A. Find the value of $f(5)$, if $f(x)$ = $x^2$ - 3.
B. If $f(x)$ = $x$ and $g(x)$ = $x^2$, then find the value of $\frac{f}{g}(x)$?

Solution:
A. Given that,
$f(x)$ = $x^2$ - 3
Now, substitute 5 in the place of $x$ i.e.
$f(5)$ = $5^2$ - 3
$f(5)$ = 22
B. Given that,
$f(x)$ = $x$
$g(x)$ = $x^2$
Now, $\frac{f}{g}(x)$ = $\frac{f(x)}{g(x)}$ where $g(x)$ ≠ 0.
$\frac{f}{g}(x)$ = $\frac{x}{x^2}$
$\frac{f}{g}(x)$ = $\frac{1}{x}$

5. A. If $h(x)$ = $x^2 + 2x$ and $j(x)$ = $\mid\frac{x}{4} + 2\mid$, then find the value of $j(h(x))$?
B. If $f(x)$ = $3x + 1$ and $g(x)$ = $\sqrt{5x}$. What is $g(f(x))$?
Solution:
A. Given that,
$h(x)$ = $x^2 + 2x$
$j(x)$ = $\mid\frac{x}{4} + 2\mid$
Initially consider $h(x)$ = $x^2 + 2x$
Substitute 4 in the place of $x$ i.e.
$h(4)$ = $4^2 + 2(4)$
$h(4)$ = 24
Now, $j(h(x))$ = $j(24)$
So, substitute 24 in the place of x in $j(x)$ = $\mid\frac{x}{4} + 2\mid$
$j(24)$ = $\mid\frac{24}{4} + 2\mid$
$j(24)$ = $\mid6 + 2\mid$
$j(24)$ = 8
Therefore, $j(h(x))$ = $j(24)$ = 8.
B. Given that,
$f(x)$ = $3x + 1$
$g(x)$ = $\sqrt{5x}$
Now, $g(f(x))$ = $g(3x + 1)$
So, substitute $3x + 1$ in the place of $x$ in $g(x)$ = $\sqrt{5x}$ i.e.
$g(f(x))$ = $\sqrt{5(3x + 1)}$
$g(f(x))$ = $\sqrt{15x + 5}$
Therefore, $g(f(x))$ = $\sqrt{15x + 5}$.