/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 If \(f(x)=x+5\) and \(g(x)=x^{2}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(f(x)=x+5\) and \(g(x)=x^{2}-3,\) find the following. $$ \begin{array}{ll}{\text { a. } f(g(0))} & {\text { b. } g(f(0))} \\ {\text { c. }} {f(g(x))} & {\text { d. } g(f(x))} \\ {\text { e. }} {f(f(-5))} & {\text { f. } g(g(2))} \\ {\text { g. }} {f(f(x))} & {\text { h. } g(g(x))}\end{array} $$

Short Answer

Expert verified
a: 2, b: 22, c: x^2 + 2, d: x^2 + 10x + 22, e: 5, f: -2, g: x + 10, h: x^4 - 6x^2 + 6

Step by step solution

01

Evaluate f(g(0))

First, find \( g(0) \) by substituting \( 0 \) into \( g(x) = x^{2} - 3 \). This gives \( g(0) = 0^{2} - 3 = -3 \). Now, substitute \( g(0) = -3 \) into \( f(x) = x + 5 \). Therefore, \( f(g(0)) = f(-3) = -3 + 5 = 2 \).
02

Evaluate g(f(0))

Find \( f(0) \) by substituting \( 0 \) into \( f(x) = x + 5 \). This gives \( f(0) = 0 + 5 = 5 \). Now, substitute \( f(0) = 5 \) into \( g(x) = x^{2} - 3 \). Thus, \( g(f(0)) = g(5) = 5^{2} - 3 = 25 - 3 = 22 \).
03

Evaluate f(g(x))

Substitute \( g(x) = x^{2} - 3 \) into \( f(x) \). This gives \( f(g(x)) = f(x^{2} - 3) = (x^{2} - 3) + 5 = x^{2} + 2 \).
04

Evaluate g(f(x))

Substitute \( f(x) = x + 5 \) into \( g(x) \). This gives \( g(f(x)) = g(x + 5) = (x + 5)^{2} - 3 = x^{2} + 10x + 25 - 3 = x^{2} + 10x + 22 \).
05

Evaluate f(f(-5))

First, find \( f(-5) \) by substituting \( -5 \) into \( f(x) = x + 5 \). This gives \( f(-5) = -5 + 5 = 0 \). Then, substitute \( f(-5) = 0 \) into \( f(x) \). Therefore, \( f(f(-5)) = f(0) = 0 + 5 = 5 \).
06

Evaluate g(g(2))

First, find \( g(2) \) by substituting \( 2 \) into \( g(x) = x^{2} - 3 \). This gives \( g(2) = 2^{2} - 3 = 4 - 3 = 1 \). Then, substitute \( g(2) = 1 \) into \( g(x) \). Therefore, \( g(g(2)) = g(1) = 1^{2} - 3 = 1 - 3 = -2 \).
07

Evaluate f(f(x))

Substitute \( f(x) = x + 5 \) into \( f(x) \). This gives \( f(f(x)) = f(x + 5) = (x + 5) + 5 = x + 10 \).
08

Evaluate g(g(x))

Substitute \( g(x) = x^{2} - 3 \) into \( g(x) \). This gives \( g(g(x)) = g(x^{2} - 3) = (x^{2} - 3)^{2} - 3 = x^{4} - 6x^{2} + 9 - 3 = x^{4} - 6x^{2} + 6 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a crucial concept in mathematics. It involves substituting a specific value into a function and calculating the result. Think of a function as a machine: you input a number, and it outputs another number based on the function's rule.
For example, if you have the function \( f(x) = x + 5 \), and you want to evaluate it at \( x = 0 \), you substitute 0 in place of \( x \). This gives \( f(0) = 0 + 5 = 5 \).
The same process applies for more complicated functions and values. Every function has a unique rule for evaluation which helps in understanding its behavior across different inputs.
Key Points:
  • Always substitute the given value into the function.
  • Simplify carefully to find the result.
  • Function evaluation helps predict how changes in input affect the outcome.
Algebraic Functions
Algebraic functions are functions that are formed by operations like addition, subtraction, multiplication, division, and raising to a power. These are confined to only polynomial functions, radical expressions, and rational functions.
Understanding algebraic functions involves recognizing their structure and behavior. Functions like \( f(x) = x + 5 \) and \( g(x) = x^2 - 3 \) are examples where basic operations shape the function's expression.

Polymeric operations can modify the shape and position of the graph of a function, making algebraic functions an essential study area for students aiming to understand calculus or more advanced topics. Remember:
  • They are defined using algebraic expressions.
  • Operations include addition, multiplication, and more complex ones like powers.
  • They are fundamental in building more complex functions seen in calculus.
Quadratic Functions
Quadratic functions are a special class of algebraic functions and are characterized by their degree, which is two. This means they include terms with \( x^2 \).
A basic representation is \( g(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic functions typically graph as parabolas, which either open upwards or downwards depending on the sign of \( a \).
In the original exercise, \( g(x) = x^2 - 3 \) represents a quadratic function. Its graph will form a parabola that opens upwards. Important attributes include the vertex, axis of symmetry, and roots, which are the solutions to \( ax^2 + bx + c = 0 \). Understanding these helps in predicting and interpreting the shape and direction of the function's graph.
  • Intercepts tell us where the function crosses the axes.
  • The vertex indicates the peak or lowest point.
  • The direction of the parabola is decided by the "\( a \)" value in \( ax^2 \).

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