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anonymous

  • 4 years ago

Can someone show me step by step how this problem is done?

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    i have an idea and notes but would like this problem clarified

  3. anonymous
    • 4 years ago
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    i know for sure its not differentiable at corners

  4. anonymous
    • 4 years ago
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    Well, all of those functions are continuous and differentiable by themselves, so your only candidates for non-differentiable points are at the places these functions overlap, or in this case, 2 and 7. However, we find that \[\lim_{x \rightarrow 7^-} f(x) = 2 - (7) = -5\] , but \[\lim_{x \rightarrow 7^+} f(x) = 7^2 + 6 = 55\] , so the function is not continuous at x=7, eliminating that option. Our only other option is x = 2, and the function approaches the same value (0) from each side, so it is continuous. In addition, because the derivatives of the functions (x-2) and (2-x) are different for all x, it follows that the slope from the left and the slope from the right at x = 2 are different, making the function non-differentiable at x=2, so the only point where the function is both continuous and non-differentiable is at x=2. Hope I explained that clearly enough!

  5. anonymous
    • 4 years ago
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    could you show me what you said using numbers and graphs rather than words. im more of a visual person when it comes to math.

  6. anonymous
    • 4 years ago
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    Sure thing, one second.

  7. anonymous
    • 4 years ago
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    ok, sweet. sorry if im being a pain.

  8. anonymous
    • 4 years ago
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    |dw:1328394320991:dw|

  9. anonymous
    • 4 years ago
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    Sorry I'm not a better artist, but basically, at x = 2, the function approaches the same value from the left and right, but at different rates, so it is continuous, but not differentiable. At x = 7, there is a massive jump in the graph, so it is not continuous nor differentiable

  10. anonymous
    • 4 years ago
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    how did you put the info given in your calculator to get this graph

  11. anonymous
    • 4 years ago
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    ?

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spraguer (Moderator)
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