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  A Question from City Tech Student

  Posted October 10, 2021|1 Reply

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  Differentiate $\displaystyle f(x) = \frac{x^{2}-3 x+2}{x^{2}}.$
  
  Answer: $\displaystyle \frac{d f}{d x} =$ ___
  

  My question:

  I don't understand why my answer isn't right

  Derivatives_-_Power_Rule | MAT1475

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  Question from a Student

  Posted October 5, 2021|1 Reply|Edit

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  $$f(x) = \left\{ \begin{array}{c l} \dfrac{x^{2}+4x-21}{-x^{2}+9}, & \, x \neq 3 \\ C, & \, x = 3 \end{array} \right.$$
  
  What value of $C$ would make $f(x)$ continuous at $x = {3}$? ___
  

  • Decimal approximations are not allowed for this problem.
    
  • Compute the exact value for $C$ and express your answer algebraically.

  
  

  My question:

  How do I do it?

  Limits_-_Continuity | MAT1475

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  Question from a Student

  Posted October 5, 2021|1 Reply|Edit

  Show WeBWorK Problem

  Continuous Functions

  
  
  A function $f(x)$ is continuous at $x=a$ if:
  

  • $a$ is in the domain of $f(x)$, i.e. $\color{red}{f(a)}$ exists
    
  • $\color{blue}{\displaystyle\lim_{x\to a} f(x)}$ exists, i.e. both one-sided limits exist and are equal
    
  • $\color{blue}{\displaystyle\lim_{x\to a} f(x)} = \color{red}{f(a)}$

  Reasons why $f(x)$ might fail to be continuous:
  $f(a)$ does not exist	$\displaystyle \lim_{x\to a}f(x)$ doesn't exist	or both exist, but don't match
  $$f(x) = (x-1)\left(\frac{x+2}{x+2}\right)$$	$$f(x) = \left\{<br />\begin{array}{ll}<br />x-1, & \, x < -2 \\ x+1, & \, x \geq -2 \end{array} \right.$$	$$f(x) = \left\{ \begin{array}{ll} (x-1)\left(\dfrac{x+2}{x+2}\right), & \, x \neq -2 \\ -1, & \, x = -2 \end{array} \right.$$

  Practice

  
  
  Determine the most accurate statement for each of the following graphs:
  

  

  

  
  
  ___ A. $f(x)$ is continuous at $x = 2$.
  
  ___ B. $f(x)$ is not continuous at $x = 2$ because $f(2)$ is undefined.
  
  ___ C. $f(x)$ is not continuous at $x = 2$ because $\displaystyle \lim_{x \to 2} f(x)$ does not exist.
  
  ___ D. $f(x)$ is not continuous at $x = 2$ because $f(2)$ and $\displaystyle \lim_{x \to 2} f(x)$ exist, but are not equal.
  
  

  
  
  ___ A. $f(x)$ is continuous at $x = 3$.
  
  ___ B. $f(x)$ is not continuous at $x = 3$ because $f(3)$ is undefined.
  
  ___ C. $f(x)$ is not continuous at $x = 3$ because $\displaystyle \lim_{x \to 3} f(x)$ does not exist.
  
  ___ D. $f(x)$ is not continuous at $x = 3$ because $f(3)$ and $\displaystyle \lim_{x \to 3} f(x)$ exist, but are not equal.
  
  

  

  

  
  
  ___ A. $f(x)$ is continuous at $x = -2$.
  
  ___ B. $f(x)$ is not continuous at $x = -2$ because $f(-2)$ is undefined.
  
  ___ C. $f(x)$ is not continuous at $x = -2$ because $\displaystyle \lim_{x \to -2} f(x)$ does not exist.
  
  ___ D. $f(x)$ is not continuous at $x = -2$ because $f(-2)$ and $\displaystyle \lim_{x \to -2} f(x)$ exist, but are not equal.
  
  

  
  
  ___ A. $f(x)$ is continuous at $x = 3$.
  
  ___ B. $f(x)$ is not continuous at $x = 3$ because $f(3)$ is undefined.
  
  ___ C. $f(x)$ is not continuous at $x = 3$ because $\displaystyle \lim_{x \to 3} f(x)$ does not exist.
  
  ___ D. $f(x)$ is not continuous at $x = 3$ because $f(3)$ and $\displaystyle \lim_{x \to 3} f(x)$ exist, but are not equal.
  
  

  
  Note: In order to get credit for this problem all answers must be correct.

  
  

  My question:

  I'm not sure how to answer this question. Is it possible that there is more than one correct statement for an image?

  Limits_-_Continuity | MAT1475

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