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

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(Created page with "<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}, &...") |
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! Foundations: | ! Foundations: | ||

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− | |A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: - | + | |A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -12%;">x_0 </math> if |

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

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! Version I: | ! Version I: | ||

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+ | |(a) For continuity, we evaluate both rules for the function at the transition point <math style="vertical-align: -5%;">x=1</math>, set the results equal, and then solve for <math style="vertical-align: 0%;">C</math>. Since we want | ||

+ | |- | ||

+ | | <math>f(1)\,=\,1\,=\,4\cdot 1^2+C,</math> | ||

+ | |- | ||

+ | |we can set <math style="vertical-align: -3%;">C=-3</math>, and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point <math style="vertical-align: -5%;">x=1</math> ). | ||

+ | (b) To test differentiability, we note that for <math style="vertical-align: -5%;">x<1</math>, | ||

+ | |- | ||

+ | | <math>f'(x)=\frac{1}{2\sqrt{x}},</math> | ||

+ | |- | ||

+ | |while for <math style="vertical-align: -10%;">x> 1</math>, | ||

+ | |- | ||

+ | | <math>f'(x)=8x.</math> | ||

+ | |- | ||

+ | |Thus | ||

+ | |- | ||

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

+ | |- | ||

+ | |but | ||

+ | |- | ||

+ | | <math>\lim_{x\rightarrow 1^+}f'(x)\,=\,8\cdot1\,=\,8.</math> | ||

+ | |- | ||

+ | |Since the left and right hand limit do not agree, the derivative does not exist at the point <math style="vertical-align: -5%;">x=1</math>.<br> | ||

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## Revision as of 13:18, 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. |

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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