Calculus/Critical points of a graph.
Expert: Paul Klarreich - 12/8/2007
QuestionHi, I have to graph the equation f(x)=x+ (32/x^2) and I am confused because I do not know how to find the critical numbers to begin the problem when there is only part of the equation in a fraction. I know that I have to do the first and 2nd derivative tests, but what should I do to find the critical number?
AnswerQuestioner: Annie
Category: Calculus
Private: No
Subject: Finding critical #
Question: Hi, I have to graph the equation f(x)=x+ (32/x^2) and I am confused because I do not know how to find the critical numbers to begin the problem when there is only part of the equation in a fraction. I know that I have to do the first and 2nd derivative tests, but what should I do to find the critical number?
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Hi, Annie,
You find the critical numbers the same way, whether there is one fractional part or 50.
Critical numbers are numbers:
1. Where f'(x) = 0 [These are called Stationary points]
2. Where f'(x) is undefined. [These are called Singular points]
3. End points of the interval, if any. [These are called, er.. End points.]
You have f(x) = x - 32x^-2 << that's the way to write it for this purpose.
f'(x) = 1 + 64x^-3
64
f'(x) = 1 + -----
x^3
Set that equal to zero:
x^3 = - 64, x = -4 is a staionary point.
Also, x = 0 is a singular point, because f'(0) is undefined.
Now:
f'(x) = 1 - 64x^-3
f''(x) = 192x^-4 = 192/x^4.
There is no inflection point.