Question

Designer functions Sketch the graph of a function \(f\) that is continuous on \((-\infty, \infty)\) and satisfies the following sets of conditions. \(f^{\prime}(x)>0,\) for all \(x\) in the domain of \(f^{\prime} ; f^{\prime}(-2)\) and \(f^{\prime}(1)\) do not exist; \(f^{\prime \prime}(0)=0\)

Short answer

Question: Sketch a graph of a function, \(f\), that is continuous everywhere and satisfies the following conditions: \(f'(x) > 0\) for all \(x\) in the domain of \(f'\), \(f'(-2)\) and \(f'(1)\) do not exist, and \(f''(0) = 0\). Describe the main features of the sketch based on the given conditions. Answer: The graph of function \(f\) is continuous everywhere, increasing except at points \(x=-2\) and \(x=1\), and has vertical tangent lines at \(x=-2\) and \(x=1\). There might be a point of inflection at \(x=0\) if the concavity changes around that point. Since we don't have explicit information about \(f(x)\), our sketch is an approximation based on the given conditions.

Step-by-step solution

Analyze given information

First, let's take a closer look at the given conditions: 1. \(f'(x) > 0\) for all \(x\) in the domain of \(f'\). 2. \(f'(-2)\) and \(f'(1)\) do not exist. 3. \(f''(0) = 0\).

Determine intervals of increasing/decreasing function

Since \(f'(x) > 0\) for all \(x\) in the domain of \(f'\), that means the function \(f(x)\) is increasing everywhere except at \(x = -2\) and \(x = 1\), where the derivative does not exist.

Determine function's concavity

As \(f''(0) = 0\), this means that there is a possible point of inflection at \(x=0\). However, we must check the sign change of \(f''(x)\) around \(x=0\) to be sure.

Create a rough sketch of the function

Based on the given information, make a rough sketch of the function: - The function is continuous everywhere. - The function is increasing everywhere except at \(x=-2\) and \(x=1\). - There might be a point of inflection at \(x=0\). We can start by drawing a continuous curve that increases from left to right. We need to make sure that at points \(x=-2\) and \(x=1\) there is a vertical tangent line (derivative does not exist). Then, we can adjust the concavity around \(x=0\) based on our analysis in the previous step to check for any point of inflection. Note: Since we are not given explicit information about \(f(x)\), we can't sketch a unique accurate graph. But our sketch should be close to what the actual function's graph would look like given the provided conditions.

Key concepts

Increasing Function

An increasing function is a key concept in calculus and signifies that as the input value (typically shown as \(x\)) increases, the output value \(f(x)\) also increases. We identify an increasing function by examining its first derivative.

• If \(f'(x) > 0\), the function is increasing at every point in its domain.
• If the sign of the derivative is not positive or undefined at certain points, this provides specific critical insights into the function's behavior.

For the function in our problem, \(f'(x) > 0\) for all \(x\), which tells us that the function is increasing everywhere, except potentially at points where the derivative does not exist, specifically \(x = -2\) and \(x = 1\). These points need special consideration, like vertical tangents or cusps, which can cause the derivative to be undefined.

Point of Inflection

A point of inflection is a point on a curve where the concavity changes. In simpler terms, it's where the curve "turns inside out" from being curved upwards (concave up) to downwards (concave down) or vice versa.

This concept becomes interesting when the second derivative \(f''(x)\) is zero. Specifically:

• If \(f''(x)\) changes sign at a particular value of \(x\), that suggests that point is a point of inflection.
• Therefore, a zero value of the second derivative at a point only indicates a potential point of inflection, not a guaranteed one.

In our problem, \(f''(0) = 0\), so there is a potential point of inflection at \(x = 0\). However, proper verification would require examining the second derivative's behavior (sign change) around this point.

Vertical Tangent Line

A vertical tangent line occurs at a point on the function where the derivative is undefined, but the function itself is still continuous. This situation typically expresses itself when the graph's direction steeply changes, such as at cusps or sharp turns.

• For our exercise, these critical points occur where \(f'(-2)\) and \(f'(1)\) do not exist.
• This means that at these points, the slope of the tangent (the derivative) becomes infinite or undefined, which reflects in the vertical appearance of the tangent line.
• Despite the undefined derivative, the function remains continuous throughout its range, including these points.

Understanding vertical tangents is crucial for thoroughly interpreting the behavior of complex, continuous functions, especially in sketching accurate graphs.

Graph Sketching

Graph sketching is a vital skill in calculus that combines understanding derivatives and function behavior to visualize the function's form. The process involves translating analytic data into a geometric representation.

When sketching the graph of any function, follow these guidelines:

• Identify intervals where the function is increasing or decreasing using the derivative.
• Locate critical points, such as where derivatives are zero or undefined.
• Determine possible points of inflection by analyzing the second derivative.

In our specific problem scenario, knowing that \(f\) is continuous and increasing almost everywhere aids significantly in drawing. Also:

• Expect vertical tangent lines at \(x=-2\) and \(x=1\) given the derivative's undefined status there.
• Consider the possible point of inflection at \(x=0\), which must be checked by the concavity change surrounding it.

The goal isn't to produce an exact graph without additional information but rather an informed and approximate representation, respecting all the fundamental behaviors outlined.