calculus, critical numbers?

Question

Find the critical numbers of the function.
(p-2) divided by ((p^2)+3)

Answer 1

y=(p-2)/(p^2+3)
y´= 1/(p^2+3)^2 * (p^2+3 -2p(p-2))
-p^2+4p+3=0 so p= ((-4+-sqrt(16+12)/(-2) = 2+-sqrt(7) which are the critical numbers

Answer 2

Critical number of a function is the value of the independent variable(in the present case p) for which either the function has the value zero or is undefined.
Critical number of the function is given by p - 2 = 0
=> p = 2
As the denominator never becomes zero, function is defined for all values of p.
So only one critical number which is 2.

Answer 3

Critical numbers are where the derivative is 0 or undefined, not the original function. Zeros in the function define asymptotes, zeros in the first derivative define local maxima and minima, and zeros in the second derivative define inflection points and concavity.

Take the derivative using the quotient rule:
([f' * g] - [f * g']) / [g^2])
then solve for p.

Answer 4

hi, you're incredible once you're saying the spinoff of f is undefined for x ? 0. the optimum question you should ask your self is, what's the area of f? that's through fact the set of severe numbers ought to be contained interior the area in accordance to the definition. So the area of f is 0? x < ?. for this reason the only severe numbers on your area are: 0, 4/9. maximum suitable.