/*! 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 38 Draw the graph of a function tha... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 38

Draw the graph of a function that is increasing on the interval [-2,0] and decreasing on the interval [0,2] .

Short Answer

Expert verified
The required graph can be represented by the piecewise function \(h(x)\): \[h(x) = \begin{cases} x^2 + 2x & \text{ if } x \in [-2, 0]\\ -x^2 + 2x & \text{ if } x \in [0, 2] \end{cases}\] which consists of an upward-facing parabola, \(f(x) = x^2 + 2x\) for \(x \in [-2, 0]\), and a downward-facing parabola, \(g(x) = -x^2 + 2x\) for \(x \in [0, 2]\). Graph the two parabolas together within their respective intervals with the vertex at the point (0,0).

Step by step solution

01

Define the upward-facing parabola for the increasing interval [-2,0]

To define an upward-facing parabola that increases in the interval [-2,0], we can use the function \(f(x) = ax^2 + bx + c\) for \(x \in [-2, 0]\). We need to find the coefficients a, b, and c such that the parabola passes through the points (-2,0) and (0,0). From the given points: 1. When x = -2, y = 0: \(0 = 4a - 2b + c\) 2. When x = 0, y = 0: \(0 = c\) Let's simplify the equation: 1. \(c = 0\) 2. \(4a - 2b = 0\) Now we need to choose values for a and b such that the conditions are satisfied. Let a = 1, then b = 2. This gives us the quadratic function: \[f(x) = x^2 + 2x\]
02

Define the downward-facing parabola for the decreasing interval [0,2]

To define a downward-facing parabola that decreases in the interval [0,2], we can use a function of the form \(g(x) = -Ax^2 + Bx + C\) for \(x \in [0, 2]\). We need to find the coefficients A, B, and C such that the parabola passes through the points (0,0) and (2,0). From the given points: 1. When x = 0, y = 0: \(0 = C\) 2. When x = 2, y = 0: \(0 = -4A + 2B + C\) Let's simplify the equation: 1. \(C = 0\) 2. \(-4A + 2B = 0\) Now we need to choose values for A and B such that the conditions are satisfied. Let A = 1, then B = 2. This gives us the quadratic function: \[g(x) = -x^2 + 2x\]
03

Combine the functions into a piecewise function

Now that we have a function for each of the specified intervals, we can combine them into a single piecewise function: \[h(x) = \begin{cases} x^2 + 2x & \text{ if } x \in [-2, 0]\\ -x^2 + 2x & \text{ if } x \in [0, 2] \end{cases}\]
04

Graph the piecewise function

To draw the graph of the function, plot the upward-facing parabola determined by f(x) in the interval [-2,0], and the downward-facing parabola determined by g(x) in the interval [0,2]. You can use a graphing tool or simply draw it on paper. Remember to include the endpoints (-2,0) and (2,0), and the vertex at the point (0,0).

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Most popular questions from this chapter

Give an example to show that the sum of two one-to-one functions is not necessarily a one-to-one function.

For each of the functions \(f\) given. (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(f^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part (c) by verifying that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I\) (recall that \(I\) is the function defined by \(I(x)=x\) ). \(f(x)=x^{2}+8\), where the domain of \(f\) equals \((0, \infty)\).

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