/*! 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 25 Sketch the slope graph of a func... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the slope graph of a function \(f\) with input \(t\) that meets these criteria: \- \(f(-2)=5\) \- the slope is positive for \(t<2\), \- the slope is negative for \(t>2\), and \(f^{\prime}(2)\) does not exist.

Short Answer

Expert verified
The graph is increasing for \(t < 2\), decreasing for \(t > 2\), with a corner at \(t = 2\).

Step by step solution

01

Understand the Conditions

We need to create a graph of a function \(f\) based on the given criteria. First, recognize that the function is defined at \(f(-2) = 5\). This means there is a point on the graph at \(t = -2\) with a height of \(5\).
02

Analyze Slope Conditions

The problem specifies that the slope is positive for \(t < 2\). This means the graph is increasing as it approaches \(t = 2\) from the left. Conversely, the slope is negative for \(t > 2\), so the graph is decreasing as it moves away from \(t = 2\).
03

Consider Slope at \(t = 2\)

We are informed that \(f' (2)\) does not exist. Typically, this means there is a discontinuity, a corner, or a vertical tangent at \(t = 2\). One common scenario leading to non-existence of a derivative is either a cusp or a corner point on the graph.
04

Sketch the Graph

Begin by plotting the point \((t, f(t)) = (-2, 5)\). Then sketch the graph so it rises toward increasing \(t\) up to just before \(t = 2\). At \(t = 2\), create a sharp corner or a cusp to transition into a decreasing slope for \(t > 2\). This ensures the slope changes from positive to negative at \(t = 2\), complying with the given conditions.

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It provides the slope of the tangent line to the function at any given point. For a given function \(f(t)\), the derivative \(f'(t)\) represents the rate of change of the function's value concerning changes in \(t\).
  • If \(f'(t) > 0\), the function is increasing (rising as you move to the right).
  • If \(f'(t) < 0\), the function is decreasing (falling as you move to the right).
  • If \(f'(t) = 0\), the function has a horizontal tangent, indicating a potential local maximum, minimum, or point of inflection.
It's important to consider points where the derivative does not exist. At these points, the function may have cusps, corners, or vertical tangents, which are characteristics the original problem alludes to at \(t = 2\). Understanding how to calculate and interpret the derivative is essential for sketching accurate slope graphs.
Continuity
Continuity in a function implies that the graph can be drawn without lifting the pencil from the paper. In mathematical terms, a function \(f(t)\) is continuous at a point \(t = a\) if the limit of \(f(t)\) as \(t\) approaches \(a\) from both sides equals \(f(a)\). This property does not guarantee the existence of a derivative.
Continuity is crucial for predicting the behavior of functions. However, at \(t = 2\), where \(f'(2)\) does not exist, the function might not be differentiable even if it is continuous. Non-differentiability does not mean non-continuity; the graph could still be continuous but not smooth.
In the context of the problem, the lack of a derivative at \(t = 2\) suggests a non-smooth transition point such as a cusp or corner, yet it does not inherently mean a break in the graph's continuity.
Cusp
A cusp is a point on the graph of a function where the direction of the curve changes sharply, leading to a nonexistent derivative due to infinite slope. It often resembles a pointed tip that distinguishes itself from smooth curves or standard corners.
In our problem, the cusp is likely occurring at \(t = 2\), as \(f'(2)\) does not exist. This is suggested by the graph's shift from a positive slope to a negative slope without a smooth transition.
Cusps create challenges for differentiation because the "sharp" change in direction causes the derivative to cease existing. However, this does not disrupt the overall graph's continuity. Recognizing cusps is crucial for accurately interpreting functions and their behavior, particularly in complex graphs.
Increasing and Decreasing Functions
Understanding whether a function is increasing or decreasing is integral to analyzing its behavior.
  • An increasing function means the graph rises as the input value \(t\) increases, indicating \(f'(t) > 0\).
  • A decreasing function occurs when the graph falls as the input value \(t\) increases, implying \(f'(t) < 0\).
In this exercise, for \(t < 2\), the function is increasing, meaning the slope graph shows a positive direction. When \(t > 2\), the function becomes decreasing, showing a negative slope graph.
These changes in the slope are essential for understanding the complete behavior of the function over its domain. Identifying increasing and decreasing intervals helps sketch a more accurate and realistic graph of the original function \(f(t)\). This insight is key to predicting the behavior of mathematical models in various applied contexts.

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Most popular questions from this chapter

Corn Crop \(\quad\) The function \(C\) gives the number of bushels of corn produced on a tract of farmland that is treated with \(f\) pounds of nitrogen per acre. a. Is it possible for \(C(90)\) to be negative? Explain. b. What are the units on \(\left.\frac{d C}{d f}\right|_{f=90}\) ? c. Is it possible for \(\left.\frac{d C}{d f}\right|_{f=90}\) to be negative? Explain.

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