# Difference between revisions of "009A Sample Final A, Problem 3"

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[[File:009A_SF_A_3.png|right|230px|frame|Both functions with constants chosen to provide continuity.]] | [[File:009A_SF_A_3.png|right|230px|frame|Both functions with constants chosen to provide continuity.]] | ||

− | <span | + | <span class="exam">3. (Version I) Consider the following function: |

<math style="vertical-align: -80%;">f(x) = \begin{cases} \sqrt{x}, & \mbox{if }x\geq 1, \\ 4x^{2}+C, & \mbox{if }x<1. \end{cases}</math> | <math style="vertical-align: -80%;">f(x) = \begin{cases} \sqrt{x}, & \mbox{if }x\geq 1, \\ 4x^{2}+C, & \mbox{if }x<1. \end{cases}</math> | ||

<br> | <br> | ||

− | <table border="0" | + | <table border="0" class="exam"> |

<tr style="vertical-align:top"> | <tr style="vertical-align:top"> | ||

<td> (a) </td> | <td> (a) </td> | ||

Line 13: | Line 13: | ||

</table> | </table> | ||

− | <span | + | <span class="exam">3. (Version II) Consider the following function: |

<math style="vertical-align: -80%;">g(x)=\begin{cases} | <math style="vertical-align: -80%;">g(x)=\begin{cases} | ||

\sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\ | \sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\ | ||

\frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1. | \frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1. | ||

\end{cases}</math> | \end{cases}</math> | ||

− | <table border="0" | + | <table border="0" class="exam"> |

<tr style="vertical-align:top"> | <tr style="vertical-align:top"> | ||

<td> (a) </td> | <td> (a) </td> | ||

Line 33: | Line 33: | ||

! Foundations: | ! Foundations: | ||

|- | |- | ||

− | |A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -14% | + | |A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -14%">x_0 </math> if |

|- | |- | ||

− | | <math>\lim_{x\rightarrow | + | | <math>\lim_{x\rightarrow x_{_0}} f(x) = f\left(x_0\right).</math> |

|- | |- | ||

− | |This can be viewed as saying the left and right hand limits exist, and are equal. For problems like these, where we are trying to find a particular value for  <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes. | + | |This can be viewed as saying the left and right hand limits exist, and are equal to the value of <math style="vertical-align: -20%">f</math> at <math style="vertical-align: -14%">x_0</math>. For problems like these, where we are trying to find a particular value for  <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes. |

|- | |- | ||

|When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledging that the derivative is a limit, and for a limit to exist it must agree from the left and the right. This means we must show the derivatives agree for both the descriptions of <math style="vertical-align: -20%;">f</math> at the transition point.<br> | |When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledging that the derivative is a limit, and for a limit to exist it must agree from the left and the right. This means we must show the derivatives agree for both the descriptions of <math style="vertical-align: -20%;">f</math> at the transition point.<br> |

## Latest revision as of 09:33, 12 April 2015

3. (Version I) Consider the following function:

(a) | Find a value of which makes continuous at |

(b) | With your choice of , is differentiable at ? Use the definition of the derivative to motivate your answer. |

3. (Version II) Consider the following function:

(a) | Find a value of which makes continuous at |

(b) | With your choice of , is differentiable at ? Use the definition of the derivative to motivate your answer. |

Foundations: |
---|

A function is continuous at a point if |

This can be viewed as saying the left and right hand limits exist, and are equal to the value of at . For problems like these, where we are trying to find a particular value for , we can just set the two descriptions of the function to be equal at the value where the definition of changes. |

When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledging that the derivative is a limit, and for a limit to exist it must agree from the left and the right. This means we must show the derivatives agree for both the descriptions of at the transition point. |

**Solution:**

Version I: |
---|

(a) For continuity, we evaluate both rules for the function at the transition point , set the results equal, and then solve for . Since we want |

we can set , and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point ).
(b) To test differentiability, we note that for , |

while for , |

Thus |

but |

Since the left and right hand limit do not agree, the derivative does not exist at the point . |

Version II: |
---|

(a) Like Version I, we begin by setting the two functions equal. We want |

so makes the function continuous. |

(b) We again consider the derivative from each side of 1. For , |

while for , |

Thus |

and |

Since the left and right hand limit do agree, the limit (which the derivative) does exist at the point , and isis differentiable at the required point. |