Use the given pair of functions to find and simplify expressions for the following functions and

Use the given pair of functions to find and simplify...

Key Concepts

Interval Notation

Interval notation is a mathematical notation used to represent a set of numbers along a continuous range. It is especially useful for describing the domains of functions, particularly when the function is not defined at certain points. Closed brackets [ ] indicate that an endpoint is included, while open brackets ( ) indicate that an endpoint is excluded.

Rational Functions and Simplification

Rational functions are functions expressed as the ratio of two polynomials. Simplifying composite functions that are rational typically involves substituting one rational expression into another, combining like terms, and reducing the expression by canceling common factors, while always checking that the cancelled factors do not introduce extraneous restrictions.

Composite Functions

Composite functions involve applying one function to the results of another. This means for two functions f and g, the composite g?f is defined as g(f(x)), and similarly, f?g is defined as f(g(x)). This concept emphasizes the order of operations where the inner function is evaluated first and its output becomes the input for the outer function.

Function Domain

The domain of a function is the set of all input values for which the function is defined. When composing functions, it is important to consider the domains of both the inner and outer functions, as the output of the inner function must lie within the domain of the outer function. Additionally, any restrictions present in the expressions, such as division by zero, must be taken into account.

Transcript

00:01 Problem we need to find composite function g of f of x is equal to g of f of x.

00:12 So g x is 2x plus 1 divide by x and we need g of f of x so we need to substitute x as f x so we have 2 times x over 2x plus 1 and this 1 is here only divide by f x that is x over 2x plus 1 this is g of f of x we also need to simplify the expressions so we have 2x over 2x plus 1 plus 1 divide by x over 2x plus 1.

01:04 If we add the numerator we have 2x plus 1 to x plus 1 in common denominator now.

01:13 So we have 2x plus 2x plus 1 divide by 2x plus 1 and this whole term is divisible by x over 2x plus 1.

01:29 So if we simplify this, we get 4x plus 1 divide by 2x plus 1 times 2x plus 1 divide by x.

01:45 So 2x plus 1 and 2x plus 1 cancels and g of f of x is 4x plus 1 divide by x this is g of f of x now for domain we have x must not be equal to zero but we must consider the domain of f x as well for the domain of f x the two x plus one must not be equal to zero so x is not equal to minus one by two so our domain will be for g of f x is minus infinity to minus 1 by 2 union minus 1 by 2 to 0 union 0 to infinity as x must not be equal to 0 and minus half.

02:54 Now next we need f of g of x so in f x we need to plug x as g x so we have g x over twice of g x plus one we have g x as 2x plus 1 so we have 2x plus 1 divide by x by 2 times 2x plus 1 1 divide by x plus 1.

03:32 If we simplify we have 2x plus 1 divide by x whole term divisible by 4x plus 2 by x and we need common denominator so multiply and divide by x so plus x by x this is in denominator.

03:55 Now let us simplify, we have 2x plus 1 divide by x.

04:03 The x goes in numerator and 4x plus 2 plus x.

04:09 That is 4x plus 2 plus x.

04:16 This.

04:18 So f of g of x is equal to the x.

04:24 The x is cancer 2x plus 1 divide by 5x plus 2 now for domain 5x plus 2 must not be equal to 0 also you need to consider the domain of g x and the domain of g x is x is not equal to 0 so x is not equal to minus 2 by 5 for f of g x and the complete domain we have is minus infinity to minus 2 by 5 or minus 0 .4...

Please give Ace some feedback