/*! 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 16 Determine whether each of the fo... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether each of the following functions is continuous and/or differentiable at \(x=1.\) $$f(x)=\left\\{\begin{array}{ll} x^{3} & \text { for } 0 \leq x<1 \\ x & \text { for } 1 \leq x \leq 2 \end{array}\right.$$

Short Answer

Expert verified
The function is continuous but not differentiable at \(x = 1\).

Step by step solution

01

Check if the function is continuous at x = 1

To determine if the function is continuous at \(x = 1\), evaluate the left-hand limit, right-hand limit, and the value of the function at \(x = 1\).
02

Evaluate the left-hand limit as x approaches 1

Since \(f(x) = x^3\) for \(0 \leq x < 1\), we need to find the limit: \[ \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} x^3 = 1^3 = 1 \]
03

Evaluate the right-hand limit as x approaches 1

Since \(f(x) = x\) for \(1 \leq x \leq 2\), we need to find the limit: \[ \lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} x = 1 \]
04

Evaluate the function at x = 1

The value of the function at \(x = 1\) is given by the second piece of the piecewise function: \( f(1) = 1 \).
05

Conclude continuity at x = 1

Since \( \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1) = 1 \), the function is continuous at \(x = 1\).
06

Check if the function is differentiable at x = 1

To determine if the function is differentiable at \(x = 1\), consider the definition of the derivative from the left and the right side.
07

Find the left-hand derivative at x = 1

The left-hand derivative of \(f(x)\) as \(x \to 1^-\) is: \[ \lim_{{h \to 0^-}} \frac{f(1+h) - f(1)}{h} = \lim_{{h \to 0^-}} \frac{(1+h)^3 - 1}{h} \]. Calculate this limit step by step.
08

Simplify the left-hand derivative

Expand \((1 + h)^3\) and simplify: \[ (1+h)^3 = 1 + 3h + 3h^2 + h^3 \]. Thus, \[ \frac{(1 + h)^3 - 1}{h} = \frac{3h + 3h^2 + h^3}{h} = 3 + 3h + h^2 \]. Taking the limit as \(h \to 0^-\), we get \[ \text{Left-hand derivative} = 3 \].
09

Find the right-hand derivative at x = 1

The right-hand derivative of \(f(x)\) as \(x \to 1^+\) is: \[ \lim_{{h \to 0^+}} \frac{f(1+h) - f(1)}{h} = \lim_{{h \to 0^+}} \frac{(1+h) - 1}{h} = \lim_{{h \to 0^+}} \frac{h}{h} = 1 \].
10

Conclude differentiability at x = 1

Since the left-hand derivative (3) does not equal the right-hand derivative (1), the function is not differentiable at \(x= 1\).

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Key Concepts

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

Piecewise Functions
Piecewise functions are defined by different expressions based on the input value, which makes them versatile for modeling situations where a process behaves differently in different scenarios. For instance, consider the function: \[f(x)=\begin{cases} x^{3} & \text{for} \ 0 \leq x < 1 \ x & \text{for} \ 1 \leq x \leq 2 \end{cases}\]Here, the function has two parts: different rules for different intervals of \(x\).
When dealing with piecewise functions, it's essential to check each segment where the rule changes, especially at the boundaries like \(x = 1\). This is crucial in analyzing continuity and differentiability, as each piece might behave differently.
Limits
Understanding limits is key to examining the behavior of functions as they approach certain points. For the given piecewise function:
  • The left-hand limit \(\lim_{{x \to 1^-}} f(x)\) evaluates the function's behavior as \(x\) approaches from values less than 1.
  • The right-hand limit \(\lim_{{x \to 1^+}} f(x)\) assesses the function as \(x\) comes from values greater than or equal to 1.

When both of these limits equal the actual function value at that point, the function is continuous. Calculations often involve substituting close values and simplifying. For our exercise, both limits equaled 1, matching \(f(1)\), ensuring the function's continuity at \(x = 1\).
Left-Hand and Right-Hand Derivatives
Derivatives measure how a function changes. To determine differentiability at a specific point, we look at the rate of change from both sides of that point.
For the left-hand derivative approaching from values \(x < 1\): \[ \lim_{{h \to 0^-}} \frac{(1+h)^3 - 1}{h} = 3 \]This is calculated by expanding and simplifying the polynomial.
The right-hand derivative for values \(x \geq 1\): \[ \lim_{{h \to 0^+}} \frac{(1 + h) - 1}{h} = 1 \]This straightforward simplification shows the function's change from the right. If these derivatives differ, the function has a 'jump' or non-smooth transition, meaning it’s not differentiable at that point. In this exercise, the left-hand and right-hand derivatives were 3 and 1 respectively, hence, the function is not differentiable at \(x = 1\).
Continuity at a Point
To determine if a function is continuous at a point, evaluate the function’s behavior as it approaches that point from both sides, as well as the actual value at the point.
The conditions for continuity at \(x = 1\) are:
  • \( \lim_{x \to 1^-} f(x) \) must exist.
  • \( \lim_{x \to 1^+} f(x) \) must exist.
  • The limit evaluations must be equal, and equal to \(f(1)\).

If any condition is unmet, the function is not continuous at that point. For our piecewise function, these conditions were met: \[\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1) = 1 \]Therefore, the function is continuous at \(x = 1\). Understanding these concepts helps in ensuring robust mathematical analysis and in identifying functions’ behavior accurately.

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Most popular questions from this chapter

Use a derivative routine to obtain the value of the derivative. Give the value to 5 decimal places. $$f^{\prime}(0), \text { where } f(x)=10^{1+x}$$

The third derivative of a function \(f(x)\) is the derivative of the second derivative \(f^{\prime \prime}(x)\) and is denoted by \(f^{\prime \prime \prime}(x) .\) Compute \(f^{\prime \prime \prime}(x)\) for the following functions: (a) \(f(x)=x^{5}-x^{4}+3 x\) (b) \(f(x)=4 x^{5 / 2}\)

Determine the value of \(a\) that makes the function \(f(x)\) continuous at \(x=0.\) $$f(x)=\left\\{\begin{array}{ll} 1 & \text { for } x \geq 0 \\ x+a & \text { for } x<0 \end{array}\right.$$

A helicopter is rising straight up in the air. Its distance from the ground \(t\) seconds after takeoff is \(s(t)\) feet, where \(s(t)=t^{2}+t.\) (a) How long will it take for the helicopter to rise 20 feet? (b) Find the velocity and the acceleration of the helicopter when it is 20 feet above the ground.

In a psychology experiment, people improved their ability to recognize common verbal and semantic information with practice. Their judgment time after \(t\) days of practice was \(f(t)=.36+.77(t-.5)^{-0.36}\) seconds. (Source: American Journal of Psychology.) (a) Display the graphs of \(f(t)\) and \(f^{\prime}(t)\) in the window \([.5,6]\) by \([-3,3]\). Use these graphs to answer the following questions. (b) What was the judgment time after 4 days of practice? (c) After how many days of practice was the judgment time about .8 second? (d) After 4 days of practice, at what rate was judgment time changing with respect to days of practice? (e) After how many days was judgment time changing at the rate of \(-.08\) second per day of practice?

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