# 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.]] | ||

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<span style="font-size:135%"><font face=Times Roman>3. (Version I) Consider the following function: | <span style="font-size:135%"><font face=Times Roman>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> |

## Revision as of 15:21, 27 March 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: |
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A function is continuous at a point if |

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 , 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 acknowledge 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: |
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(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: |
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(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 .is |