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January 24th, 2019, 02:48 AM   #1
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The answer in my book is 5 but if we put 5 instead of x all of them are differentiable
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January 24th, 2019, 03:02 AM   #2
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$|x^2 - 4|$ isn't differentiable at ±2 (two values of $x$), $\text{sgn}(x^3 + x)$ isn't differentiable at 0, etc.
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January 24th, 2019, 03:10 AM   #3
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You are misreading the question. It is not asking for the actual values of x where f(x) is not differentiable. It is asking how many such values there are.
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January 24th, 2019, 04:59 AM   #4
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Unrelated comment: This is truly a terrible question. Whoever assigned it isn't paying attention. The only reasonable answer is 3, not 5. It only makes sense to talk about differentiability of a function on its domain. Functions don't "see" anything that isn't in their domain. Both of the following sentences make an equal amount of sense.

1. $f$ is not differentiable at $x = \frac{1}{\sqrt{2}}$.

2. $f$ is not differentiable at $x = $ 1998 Honda Accord.
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Last edited by skipjack; January 24th, 2019 at 11:52 AM.
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January 24th, 2019, 01:28 PM   #5
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$x^2-4=0$, 2 values.
$x^3+x=0$, 1 real value.
$2x^2-1=0$, 2 values.
Total 5 (real) values.
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January 24th, 2019, 02:06 PM   #6
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It is enough to know how to find the derivative , then the domain can be found easily .
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