Saturday

January 31, 2015

January 31, 2015

Posted by **Jen** on Thursday, November 9, 2006 at 1:20pm.

1, f(x) = xe^x

2. f(x) = x^(3/5)

3. f(x) = x/(x-2)

TIA

1) R

2) It's the interfal from zero to infinity for f but for f' it is this inteval with zero excluded, because unlike f, f' is singular at zero.

3) Here only the point 2 has to be excluded from the set of all reals.

Thank you. But how do we get this?

The domain is the subset of the real numbers for which the function is defined. You can turn this aroiund and ask if there are any values that you cannot insert for x.

In case of the first function, there are no values that you cannot substitute for x.

In case of the second function, you are dealing with a fractional power. Fractional powers are not defined for negative numbers. If you differentiate this function the power of x becomes negative as well, and negative powers are not defined for zero.

In the last problem, you can see that the function has a singularity at x = 2.

Thanks a lot. Got it. :)

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