#### Chapter 41

5 Steps - 3 Clicks

### 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:

Given

$$f(x)$$ = -3$$x^2$$ – 12$$x$$ – 1

Consider the vertex formula with $$a$$ = -3 and b = -12

$$h$$ = $$\frac{-b}{2a}$$

= $$\frac{-12}{2(-3)}$$

= -2

$$k$$ = $$f(h)$$ = $$f(-2)$$

= -3$$(-2)^2$$ – 12(-2) – 1

= 11

Therefore, the graph is as shown below:

It is a parabola, opening downward, with vertex as $$x$$ = -2.

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, $$g(f(x))$$ = $$\sqrt{15x + 5}$$.