Function notation
Function notation In mathematics you will encounter the following very often : f(x) = . . . . or g(x) = . . . . or h(x) = . . . . We are going to spend some time on this type of notation (way of writing) to understand it well, and be able to use it properly. You know what the following means : y = 2x – 1 y = x² - 3x + 4 3 y x It is all very well, when working with them individually, to feed different x values into each to determine their outputs (y values). You will, however, encounter continuity problems when working with all of them at the same time, for instance when drawing it on the same axes-diagram. It is therefore necessary to distinguish between different functions when they are being used simultaneously. In cases like this the following notation is then used f(x) = 2x – 1
Functions
Let us just pause here for a moment to define what everything means: the f is the name of the function. Like you have a name, each of your friends has a name, the function has a name, and it is “ f ” In the function itself the independent variable is “x “, and it is therefore listed inside the brackets next to the function’s name: f ( x ) o You will NEVER encounter the following : f(y) = 2x – 1 o In advanced mathematics you may work with the following: f(x;y)=2xy + y² o Take note, then, that the variable which should, and will, be used as the input will be listed alongside the name of the function. If we then look at the previous three functions we can write them as follows : f(x) = 2x – 1 g(x) = x² - 3x + 4 h( x )
3 x
where the “ f “, “ g “ and “ h “ indicates each one’s name
Functions
You can expect to encounter the following types of questions: 1) If h(x) = 2x + 5, calculate 1) h(3) 2) h( – 3) 3) h(0) 4) h(a) 2)
Determine x if g(x) = 3x + 4 and g(x) = – 5
To solve these questions we do the following : 1) In h(x) = 2x + 5 we then now that the function’s name is h, and x is the input variable 1) Thus, if there is written h(3) we replace all the x’s in the function with 3: h(3) = 2(3) + 5 =6+5 = 11 2) h( –3) = 2(– 3) + 5 = –6+5 = –1 A hint that you can remember is to rewrite the whole function but use empty brackets wherever there is a “x”. If you are finished, and you are certain that all the exponents and coefficients are in the right places as per the original function, you can just fill in the input value into all the brackets, and simplify Example : 3) h(0) = 2( ) + 5 = 2(0) + 5 =5 Functions
4) h(a) = 2 ( ) + 5 =2(a)+5 = 2a + 5 2)
This question is exactly the opposite of question 1. In this question the value of g(x) is given, and we are required to find the specific input that produced this output. IN THIS CASE YOU WILL NOT REPLACE THE x’s WITH THE GIVEN VALUE, BUT RATHER PUT THE WHOLE FUNCTION EQUAL TO IT Thus : g(x) = 3x + 4 then the function is 3x + 4 Further it is given that g(x) = – 5 3x + 4 = – 5 3x = – 5 – 4 3x = – 9 x= –3
You could also come across the following types of questions: g : x 3x² + 5 1) Write down an algebraic equation for g(x) 2) Calculate g(2) 3) Show that g(a + b) g(a) + g(b) 1) 2)
We need to write this only as g(x) = 3x² + 5 g(2) = 3( )² + 5 = 3(2)² + 5 = 12 + 5 = 17 Functions
3)
First we do the left side, then the right side, and then we compare them to see if they are the same: g(a + b) = 3(a + b)² + 5 = 3(a² + 2ab + b²) + 5 = 3a² + 6ab + 3b² + 5 g(a) + g(b)
= = =
[3(a)² + 5] + [ 3(b)² + 5] 3a² + 5 + 3b² + 5 3a² + 3b² + 10
It is therefore clear that the Left side is not equal to the Right side. This implies for instance that g(3) + g(7) is not equal to g(10)
Functions
Functions
Let us just pause here for a moment to define what everything means: the f is the name of the function. Like you have a name, each of your friends has a name, the function has a name, and it is “ f ” In the function itself the independent variable is “x “, and it is therefore listed inside the brackets next to the function’s name: f ( x ) o You will NEVER encounter the following : f(y) = 2x – 1 o In advanced mathematics you may work with the following: f(x;y)=2xy + y² o Take note, then, that the variable which should, and will, be used as the input will be listed alongside the name of the function. If we then look at the previous three functions we can write them as follows : f(x) = 2x – 1 g(x) = x² - 3x + 4 h( x )
3 x
where the “ f “, “ g “ and “ h “ indicates each one’s name
Functions
You can expect to encounter the following types of questions: 1) If h(x) = 2x + 5, calculate 1) h(3) 2) h( – 3) 3) h(0) 4) h(a) 2)
Determine x if g(x) = 3x + 4 and g(x) = – 5
To solve these questions we do the following : 1) In h(x) = 2x + 5 we then now that the function’s name is h, and x is the input variable 1) Thus, if there is written h(3) we replace all the x’s in the function with 3: h(3) = 2(3) + 5 =6+5 = 11 2) h( –3) = 2(– 3) + 5 = –6+5 = –1 A hint that you can remember is to rewrite the whole function but use empty brackets wherever there is a “x”. If you are finished, and you are certain that all the exponents and coefficients are in the right places as per the original function, you can just fill in the input value into all the brackets, and simplify Example : 3) h(0) = 2( ) + 5 = 2(0) + 5 =5 Functions
4) h(a) = 2 ( ) + 5 =2(a)+5 = 2a + 5 2)
This question is exactly the opposite of question 1. In this question the value of g(x) is given, and we are required to find the specific input that produced this output. IN THIS CASE YOU WILL NOT REPLACE THE x’s WITH THE GIVEN VALUE, BUT RATHER PUT THE WHOLE FUNCTION EQUAL TO IT Thus : g(x) = 3x + 4 then the function is 3x + 4 Further it is given that g(x) = – 5 3x + 4 = – 5 3x = – 5 – 4 3x = – 9 x= –3
You could also come across the following types of questions: g : x 3x² + 5 1) Write down an algebraic equation for g(x) 2) Calculate g(2) 3) Show that g(a + b) g(a) + g(b) 1) 2)
We need to write this only as g(x) = 3x² + 5 g(2) = 3( )² + 5 = 3(2)² + 5 = 12 + 5 = 17 Functions
3)
First we do the left side, then the right side, and then we compare them to see if they are the same: g(a + b) = 3(a + b)² + 5 = 3(a² + 2ab + b²) + 5 = 3a² + 6ab + 3b² + 5 g(a) + g(b)
= = =
[3(a)² + 5] + [ 3(b)² + 5] 3a² + 5 + 3b² + 5 3a² + 3b² + 10
It is therefore clear that the Left side is not equal to the Right side. This implies for instance that g(3) + g(7) is not equal to g(10)
Functions
