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

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| <math>f(1)\,=\,\sqrt{1^2+3}\,=\,2\,=\,\frac{\,\,1^2}{4}+C,</math> | | <math>f(1)\,=\,\sqrt{1^2+3}\,=\,2\,=\,\frac{\,\,1^2}{4}+C,</math> | ||

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− | |so <math style="vertical-align: -24%;">C= | + | |so <math style="vertical-align: -24%;">C=7/4</math> makes the function continuous. |

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|(b) We again consider the derivative from each side of 1. For <math style="vertical-align: -2%;">x>1</math>, | |(b) We again consider the derivative from each side of 1. For <math style="vertical-align: -2%;">x>1</math>, | ||

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| <math>\lim_{x\rightarrow 1^-}f'(x)\,=\,\frac{1}{2}.</math> | | <math>\lim_{x\rightarrow 1^-}f'(x)\,=\,\frac{1}{2}.</math> | ||

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− | |Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math> | + | |Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math>, and <math style="vertical-align: -13%;">g(x)</math><br> is differentiable at the required point. |

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[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']] |

## Revision as of 11:51, 28 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 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: |
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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 , and isis differentiable at the required point. |