**Lines and Circle- Basic Formula in Lines**

a) Slope of a line (m) from starting point(a,b) to end point (x,y) is; m=(y-b)/(x-a)

b) Point-slope form:The line with the point (x_{1},y_{1}) having slope m is given by; (x-x_{1}) =m(y-y_{1})

c) Slope-Intercept Form: If m is the slope of a line and c be the y-intercept, then equation of line is; y=mx+c.

d) Two lines with having slopes m and n are parallel if m=n.

e) Two lines with slopes m and n are perpendicular if mn = -1.

f) Distance Formula: Distance between the two points (x_{1},x_{2}) and(y_{1},y_{2 }) Problems is

e) Circle with centre (h,k) and radius r is (x-h)^{2}-(y-k)^{2}=r^{2}

**2. Function (Domain and Range):**

A function takes the elements of a domain set to range set, given as;

F: XY, where X is the domain and Y is the range.

Let X= { a,b,c,d} and Y= {1,2,3,4,5}

** **

Here, Domain = {a,b,c,d} and Range={1,2,3,5}Here, f(a) =2, f(b)=1 ,f(c) = 3 and f(d) =5

We can also write it as a relation {(a,1),(b,1),(c,3),(d,5)}

Also, Codomain = {1,2,3,4,5}.

Also, Range contained Codomain.

Question1. Find the range and domain of the function.

{(1,5),(2,0),(-1,5),(3,2),(-2,-2)}

Solution: Domain={-2,-1,1,2,3} and range ={-2,0,2,5}.

3. Limits

Any real-valued function f(x) is said to be close to some real number M if the variable x approaches to some real number b.

It is written as ,

** Lim _{x → b }f(x) =M**

**There are two types of limits:**

- LHL: LHL means left hand limit which is applied to f(x) only when x approaches to b from left side. It is given as:

**Lim _{x → b+}f(x)=M**

- RHL: RHL means right hand limit which is applied to f(x) only when x approaches to b from right side. It is given as:

**Lim _{x → b-}f(x) =M**

**Note**: A limit is said to exit if and only if LHL is equals to RHL.

Or, ** Lim _{x b }f(x) =M ↔ Lim _{x b+}f(x) = Lim _{x b-}f(x)**

**Example1**: Find Lim_{x → 2}f(x) where

f(x) = 2x^{2}-4x, x<0

2x^{2}+4x, x>0

is limit exit or not ?

**Solution: **we will find LHL and RHL and check whether the two equal or not.

**Left Hand Limit is 2 and it will be done this way:**

: Lim_{x → 2-}f(x) = Lim_{x → 2 }(2x^{2}-4x) = 8-8 =0

**Right Hand Limit is**

: Lim _{x → 2}+ f(x) =Lim_{x → 2}(2x^{2}+4x) =8+8=16

So, we get that: LHL = RHL

Thus, Lim_{x2} f(x) here, doesn’t exit.

**Continuity**

A function f is said to be continuous if Lim_{x → b}f(x) = f(b)

**Discontinuity: **Not all functions are continuous, some can be discontinuous. There are three types of discontinuity:

**Infinite Discontinuity****:**If either LHL is infinite or, RHL is infinite or f(a) is infinite, then f(x) is said to be of infinite discontinuity.**Jump Discontinuity****:**When both LHL and RHL exit but LHL = RHL.**Removable Discontinues****:**When both LHL and RHL exit and LHL=RHL = f(a). We can take the value of f(a) matching with limits and can remove the discontinuity. That is why; this is known as term: removable discontinuity.

**Here is the next example:**

**Example 2: (a) **f(x)= 1/x-1 at x=1

f(x)=1/1-1= 1/0= Infinite

so, f(x) has infinite discontinuity at x= 1

**(b) **f(x)= 1, x<0

-1,x>0

Without loss of generality, LHL= 1 and RHL= -1

So, it would be jump discontinuity.

**(C) **f(x)= (x^{2}-4)/(x-2) at x=2 .

It would be removable singularity.

**Differentiability **

The derivative of a function f(x) is considered as the slope of tangent line passing through the given point. Like limits, in derivatives we also have LHD and RHD.

The differentiablity of any function,let us take f(x) at any particular point p is find as below:

f’(x)= Lim_{xp→}(f(x)- f(p))/ (x-p)

LHD = Lim_{x → p-}f(x)

Put x=c-h and take h → 0,

So that, when h → o, x p- → = Lim_{h → 0 }(f(p-h)-f(p))/(-h)

RHD= Lim_{x → c+}(f(x)-f(p))/(x-p)

Put x= c+h and take h → 0

So that, when h → o, x → p+

=Lim_{h→ 0}(f(p+h)-f(p))/h

Boost your grade by hiring an online calculus tutor!

**Happy Studying!**