Clarifying piecewise function help! (Calculus) For my homework the question asks
ID: 2838899 • Letter: C
Question
Clarifying piecewise function help!
(Calculus)
For my homework the question asks to find the critical numbers of the following piecewise function.
f(x) = 2-x^2 when 1 < x < 3 and 2-3x when 3 < x < 5 and the interval is [1,5].
I had some questions about solving this problem. I know that I have to find places where the derivative is equal to zero or does not exist. So the derivative of f(x) is -2x when 1 < x < 3 and -3 when 3 < x < 5.
I am slightly confused about the next step. Do I set all parts of the function equal to zero? If so I get -3=0 which is not possible, and x=0 which is not in the domain. This suggests that there are no critical numbers. But if the derivative is never equal to 3, does that mean there is a critical number at 3 because the derivative does not exist there, or no because it is not in the domain? Please help clarify this question, thanks!
Explanation / Answer
f(x) = 2x2 for 1<x<3
f(x) = 2-3x for 3<x<5
f'(x) = 4x for the first interval.
=-3 for the second interval.
Equate f'(x) to 0 for the first interval.
x =0 is not in [1,5]
and second interval f'(x) never equals 0.
Thus no critical point in the interval for the given f(x).