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Conditions & Steps for Solving Functions!!!!!

Functions are defined as the values in which some input has particular output depending upon it types. A function is types of relation between sets.

Let A and B be any two non empty sets such that every element has one or more than one image but image cannot have more than one element.

Condition of a function

Set A and Set B should be non-empty.

In a function, a particular input is given to get a particular output. So, A function f: A->B denotes that f is a function from A to B, where A is a domain and B is a co-domain.

  • For an element, a, which belongs to A, aϵA, a unique element b, bϵB is there such that (a,b)ϵ f.

The unique element b to which f relates a, is denoted by f(a) and is called f of a, or the value of f at a, or the image of a under f.

  • The range of (image of a under f)
  • It is the set of all values of f(x) taken together.
  • Range of f = { y ϵ Y | y = f (x), for some x in X}

A real-valued function has either P or any one of its subsets as its range. Further, if its domain is also either P or a subset of P, it is called a real function.

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Representation of a Function!

A function is generally represented as f(x).

Let f(x) = x2

It will be pronounced as “ f of x is equal to x square.”

Function is represented in the form of g( ), f( ), t( )…etc

Steps for solving functions

Example 1: Find the output of function

                     g(t)= 6t2 + 5t +9

                     At t = 1

Solution:  The given function is g(t) = 6t2 + 5t +9

                    At t =1,

                    6(1)2 + 5(1) +9 = 20

Example 2: Find f(4x-1) for f(x)= -x2+6x-11

Solution:  f (4x−1) = − (4x−1)2+6(4x−1) − 11

                               = −16x2+32x−18

This one is not much different from the previous part. All we did was change the equation that we were plugging into the function.

Example 3: Determine all the roots of f(t)  =  9t3-18t2+6t

Solutions: Let’s solve,

                  9t3−18t2+6t = 0

 First we should factorize the equation,

                  3t (3t2−6t+2) =0

If products of two things are zero then one or both of them had to be zero. That means-

                      3t = 0,   OR

             3t2-6t+2=0

From the first it’s clear that one of the roots must then be t =0. To get the remaining roots we will need to use the quadratic formula on the second equation. Doing this gives,

The whole list of roots of this function is –

   

Note we didn’t use the final form for the roots from the quadratic. This is usually where we’ll stop with the simplification for these kinds of roots. Also note that, for the sake of the practice, we broke up the compact form for the two roots of the quadratic. You will need to be able to do this so make sure that you can.