Composite functions are ones in which one function is wrapped around another. When two functions, f, and g, are joined to create a new function, h, with the formula h(x) = g(f(x)), this is known as the composition of a function in mathematics. It signifies that the function g is applied to the function x in this case. Function composition is replacing the first function’s independent variable with the expression representing the dependent variable of the second function.
Let’s say there are two functions, f: A → B and g: B→ C. The equation g ∘ f (x) = g (f (x)), ∀ x ∈ A yields the function g ∘ f: A → C. defines the composite of f and g, denoted by g f. The domain and function composition symbol with an example.
For example, if f(x) = 3x+1 and g(x) = x2, then f(g(x)) = f(x2) = 3x2+1.
If we invert the function operation, g(f(x)) = g(3x+1) = (3x+1)2
There are a few more properties:
The composition of functions is represented algebraically and graphically. Solving a composite function in mathematics entails obtaining the composition of two functions. The composition of a function is indicated by a tiny circle (∘).
Follow the steps below to learn how to solve the given composite function algebraically.
Step 1: Begin by rewriting the given composition in a new way.
Consider the equations f(x) = x2 and g(x) = 3x.
Now, f[g(x)] can be expressed as (f ∘ g) (x).
Step 2: Using the individual functions as a guide, replace the variable x in the outside function with the variable x in the inside function.
In other words, Since g(x) = 3x, (f ∘ g)(x) = f(3x).
Step 3: Finally, simplify the function you’ve created.
Since f(x) = x2, (f ∘ g)(x) = f(3x) = (3x)2=9×2
Let the function f defined by f(x)=2x+3 and the function g defined by g(x)=x2.
The compound (f ∘ g) is calculated as follows:
(f ∘ g) (x)=f(g(x))
=f(x2)
=2(x2) + 3
=2x2+ 3
There is no restriction to put here, so the domain of this composition is R.
(g∘f)(x) =g(f(x))
=g(2x+3)
=(2x+3)2
=(2x+3) (2x+3)
=4x2+12x+9
There are no restrictions to put here, so the domain of this composition is
R.
Let the function f defined by f(x)=1+x2 and the function defined by g(x)=√x.
The compound (f ∘ g)(x) is calculated as follows:
(f∘g)(x) = f(g(x))
=f(√x)
=1+(√x)2
=1+x
Here, the domain is the set of positive real numbers. In fact, under the square root, we can only put positive numbers.
The compound (g ∘ f) (x) is calculated as follows:
(g ∘ f) (x)=g(f(x))
=g(1+x2)
=√1+x2
The set of real numbers makes up the function’s domain. The function 1+x2 is always positive; therefore, the square root function is always well-defined.
An inverse function, also known as anti-function, is a function that “inverts” another function: if the function f applied to an input x yields a result of y, then using its inverse function g to y yields the result x, i.e., g ( y ) = x if and only if f ( x ) = y The inverse function of f is also noted as f – 1
For example, consider the real-valued function of a real variable given by f ( x ) = 2x − 3. Assuming that this is a step-by-step process (multiplying x by 2 and then subtracting 3 from the result), we would need to undo each step in reverse order to get x from the output value, let’s say y. In this case, that means adding 3 to y, then dividing the outcome by 2. This inverse function would be represented by, in functional notation:
g(y) = \frac{(x + 7)}{5}
With y = 2x −3, we have f(x) = y and g (y) = x.
Let f be a function with the set X as its domain and the set Y as its codomain. Then f is invertible if a function g of domain Y and codomain X exists and has the property:
f ( x )= y⇔g ( y ) = x .
If f is invertible, only one function g exists that satisfies this characteristic, making the function g unique. The ranges of g and f also follow that they are equal to their respective codomains. The function g is the inverse of f and is typically represented by f -1.
In other words, a function is regarded as binary if and only if its inverse is a function on the codomain Y, in which case the inverse relation is the inverse function.
Use the variation table to plot the curve representing the inverse function, then the coordinates of a few points on the curve are determined. The results can be collected in a table.
x –2 –1 –0.5 0.5 1 2 4
1/x –0.5 –1 –2 2 1 0.5 0.25
We thus obtain the following graphical representation:
The hyperbola passes in particular through the points:
– A(1,1), B(0.5,2), C(2,0.5);
– A(-1,-1), B(-0.5,-2), C(-2,-0.5).
Points A and A’ are symmetrical to the center of symmetry O. The same applies to points B and B,’ and C and C.’ O is the middle of the segments [AA’], [BB’], and [CC’].
Consider the function f(x) to find the inverse function f-1 (x).
Substitute y for f(x).
Replace all x’s with y’s and vice versa.
For y, solve the equation generated in step 2.
Substitute f-1 for y. (x).
This method can be used to determine the inverse for the vast majority of functions.
Example:
Find the inverse for f(x) = x2 + 1.
Solution
Follow the same steps to find out the inverse of this function.
f(x) = x2 + 1
⇒ y = x2 + 1
Replacing x with y and vice-versa
⇒x = y2 + 1
⇒ y = √(x^2-1)
Thus, f-1(x) = √(x^2-1)
Inverse returns the input. It’s critical to double-check whether the calculated inverse is correct. We leverage the property of function composition to accomplish this. Let’s say f(x) is a function, and g(x) is the inverse of it. The objective is to ensure that the estimated inverse function g is correct (x). The two requirements for proving that g(x) is the inverse of f(x) are as follows:
For all x in g’s domain, f(g(x)) = x (x).
For all x in f’s domain, g(f(x)) = x (x).
Because f(x) and g(x) are inverses, any order in which they are combined produces a function that returns input for input. The identity function is the name for this function.
Example
Find the inverse of the following function and verify it with the abovementioned properties. f(x) = 2x + 5
Solution
Following the steps
f(x) = 2x + 5
Replace f(x) with y
y= 2x + 5
Replace x with y and vice versa
x= 2x + 5
Solving the equation for y
y = \frac{5-x}{x}
Let the inverse be g(x)=\frac{5-x}{2} and f(x)= 2x + 5
f(g(x))
f(\frac{x-5}{2})
2\left( \frac{x-5}{2} \right) + 5
=x – 5 + 5
=x
g(f(x))
g(2x + 5)
=(((2x+5)-5)/2)
=x
Therefore, it is verified.
For f(x) = 2x + 3 and g(x) = -x2 + 1, find the composite function defined by (f o g)(x)
Solution
(f o g)(x) = f(g(x))
= 2 (g(x)) + 3
= 2( -x2 + 1 ) + 3
= – 2 x2 + 5
Functions f and g are given by
f(x) = √(x + 3) and g(x) = ln (1 – x 2)
Find the composite function defined by (g o f)(x) and describe its domain.
Solution
(g o f)(x) = g(f(x))
= ln (1 – f(x) 2)
= ln (1 – √(x + 3) 2)
= ln (1 – (x + 3))
= ln (- x – 2)