/*! 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 11 Let \(f\) be the function define... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(f\) be the function defined by $$ f(x)=\left\\{\begin{array}{ll} x^{2}+1 & \text { if } x \leq 0 \\ \sqrt{x} & \text { if } x>0 \end{array}\right. $$ Find \(f(-2), f(0)\), and \(f(1)\).

Short Answer

Expert verified
The short answer to the problem is: \(f(-2) = 5\), \(f(0) = 1\), and \(f(1) = 1\).

Step by step solution

01

Identify the correct function definition for \(f(-2)\)

As \(-2\) is less than or equal to \(0\), we will use the first part of the function definition to find \(f(-2)\). That is, we use the expression \(x^{2}+1\) when \(x = -2\).
02

Compute \(f(-2)\) using the chosen expression

Plug \(x = -2\) into the expression \(x^{2}+1\): $$ f(-2)=(-2)^{2}+1=4+1=5 $$ So, \(f(-2) = 5\).
03

Identify the correct function definition for \(f(0)\)

As \(0\) is less than or equal to \(0\), we will use the first part of the function definition to find \(f(0)\). That is, we use the expression \(x^{2}+1\) when \(x = 0\).
04

Compute \(f(0)\) using the chosen expression

Plug \(x = 0\) into the expression \(x^{2}+1\): $$ f(0) = (0)^{2}+1=0+1=1 $$ So, \(f(0) = 1\).
05

Identify the correct function definition for \(f(1)\)

As \(1\) is greater than \(0\), we will use the second part of the function definition to find \(f(1)\). That is, we use the expression \(\sqrt{x}\) when \(x = 1\).
06

Compute \(f(1)\) using the chosen expression

Plug \(x = 1\) into the expression \(\sqrt{x}\): $$ f(1) = \sqrt{1}=1 $$ So, \(f(1) = 1\). To summarize, we have found the following values for the given function: - \(f(-2) = 5\) - \(f(0) = 1\) - \(f(1) = 1\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Function Evaluation
When working with functions, an essential skill is function evaluation, which involves determining the output of a function given a specific input. In the context of piecewise functions, the process requires identifying which part of the function applies to the input value. This means carefully analyzing the conditions under which different expressions of the function are valid.

The example provided presents a piecewise function with two different expressions: a quadratic function used when the input is less than or equal to zero, and a square root function used when the input is greater than zero. Evaluating this function, like in the steps to find f(-2), f(0), and f(1), demands attention to the 'if' conditions to determine which expression to use for computation. It's like choosing the correct tool based on the task at hand—pick the quadratic tool for inputs less than or equal to zero and the square root tool for inputs greater than zero.

Understanding how to evaluate piecewise functions is critical because they can model a wide range of real-world scenarios, helping us to understand different behaviors depending on the given conditions. The improved way to approach function evaluation is by creating a number line to visualize where the input value falls relative to the conditions provided.
Square Root Function
A square root function is a type of function that features the square root of the variable, expressed as \( f(x) = \sqrt{x} \) for all non-negative values of \( x \). These functions are a key part of algebra and appear in various mathematical contexts. The principal square root is always non-negative, ensuring the function is well-defined for all non-negative inputs.

One of the characteristics of the square root function is that it increases gradually; for large values of \( x \) the rate of increase slows down. The function is not defined for negative values of \( x \) as the square root of a negative number is not a real number (unless one is working with complex numbers). This characteristic is illustrated in the function from the exercise for inputs greater than zero, such as \( f(1) \).

For students dealing with square root functions, visual aids such as graphs can be particularly helpful to understand the shape and behavior of the function across its domain.
Quadratic Function
Up next is the quadratic function, a staple in mathematics with the general form \( f(x) = ax^2 + bx + c \) where \( a \) is not zero. Quadratics are famous for their curved graphs, known as parabolas, and appear in various applications like projectile motion and optics.

The quadratic function is defined for all real values of \( x \). When you input a value into a quadratic function, you square it, multiply it by \( a \) if \( a \) is not equal to 1, add it to the product of \( b \) and \( x \) and finally add \( c \) to get the output. In the exercise's example, the expression \( x^2 + 1 \) is a simplified quadratic function, where \( a = 1 \) and \( b = 0 \), exhibiting a parabola that opens upwards.

For complex expressions or larger numbers, it's beneficial to use graphing tools or software to visualize the quadratic function. This can help students see the symmetry and the vertex, which are integral parts of understanding the properties of quadratic functions. Always make sure to look at the coefficient of the squared term to predict the direction in which the parabola opens—upwards if positive, and downwards if negative.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

AutoTime, a manufacturer of 24 -hr variable timers, has a monthly fixed cost of \(\$ 48,000\) and a production cost of \(\$ 8\) for each timer manufactured. The timers sell for \(\$ 14\) each. a. What is the cost function? b. What is the revenue function? c. What is the profit function? d. Compute the profit (loss) corresponding to production levels of 4000,6000 , and 10,000 timers, respectively.

The estimated monthly profit realizable by the Cannon Precision Instruments Corporation for manufacturing and selling \(x\) units of its model \(\mathrm{M} 1\) cameras is $$ P(x)=-0.04 x^{2}+240 x-10,000 $$ dollars. Determine how many cameras Cannon should produce per month in order to maximize its profits.

Find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=-4 x^{2}+4 x+3\)

For years, automobile manufacturers had a monopoly on the replacement-parts market, particularly for sheet metal parts such as fenders, doors, and hoods, the parts most often damaged in a crash. Beginning in the late \(1970 \mathrm{~s}\), however, competition appeared on the scene. In a report conducted by an insurance company to study the effects of the competition, the price of an OEM (original equipment manufacturer) fender for a particular 1983 model car was found to be $$ f(t)=\frac{110}{\frac{1}{2} t+1} \quad(0 \leq t \leq 2) $$ where \(f(t)\) is measured in dollars and \(t\) is in years. Over the same period of time, the price of a non-OEM fender for the car was found to be $$ g(t)=26\left(\frac{1}{4} t^{2}-1\right)^{2}+52 \quad(0 \leq t \leq 2) $$ where \(g(t)\) is also measured in dollars. Find a function \(h(t)\) that gives the difference in price between an OEM fender and a non-OEM fender. Compute \(h(0), h(1)\), and \(h(2)\). What does the result of your computation seem to say about the price gap between OEM and non-OEM fenders over the 2 yr?

BREAK-EvEN ANALYsIS A division of Carter Enterprises produces "Personal Income Tax" diaries. Each diary sells for \(\$ 8\). The monthly fixed costs incurred by the division are \(\$ 25,000\), and the variable cost of producing each diary is \(\$ 3\). a. Find the break-even point for the division. b. What should be the level of sales in order for the division to realize a \(15 \%\) profit over the cost of making the

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.