Calculus critical numbers question, help?

h(p) = (p - 3)/(p² + 2))

U = p - 3

U ' - 1

V = p² + 2

V ' = 2p

dy/dp = (VU ' - UV ') / V²

dy/dp = [(p² + 2) - (p - 3)(2p)] / (p² + 2)²

dy/dp = [p² + 2 - (2p² - 6p)] / (p² + 2)²

dy/dp = (p² + 2 - 2p² + 6p) / (p² + 2)²

dy/dp = (- p² + 6p + 2) / (p² + 2)²

Critical Nos.: dy/dp = 0:

(p² + 2)² = 0 would cause division by 0, therefore,

- p² + 6p + 2 = 0

p² - 6p - 2 = 0

p² - 6p = 2

p² - 6p + 9 = 2 + 9

(p - 3)² = 11

p - 3 = √11

p - 3 = ± 3.317

p = ± 3.317 + 3

p = 3.317 + 3

p = 6.317

p = - 3.317 + 3

p = - 0.317

Critical Nos.: p{- 0.317, 6.317}

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To determine critical points, sub the critical nos. for p in the original equation and solve for h(p), or, if you prefer, y.

I'm going to assume you know the quotient rule. So take the derivative of the function, and you have h'(p)=((p^2+2)(1)-(p-3)(2p))/(p^2+2)^2=p^2+2-2p^2+6p=-p^2+6p+2 (Low dee high minus high dee low all over the square of what's below).

To find critical numbers, set the derivative equal to zero. So in this case, you can disregard the denominator, and you'll have -p^2+6p+2=0. The critical numbers will be the x-intercepts of that function. So the critical numbers are x=-0.317 and x=6.317. Of course the numbers are rounded off, but that's okay. A critical number is also a number that makes the denominator of a function 0, but nothing will make the denominator 0 in this problem.

Take derivative. Set equal to 0 to find values of p that are "critical".

Take 2nd derivative. Set equal to 0. Input critical p values to see if 2nd derivative is positive, negative or 0. If positive, then it's a maximum; if negative, it's a minimum; if 0, it's an inflection point.

I assume you can take it from here. After all, it's trivial to calculate derivatives.

hi, you're maximum suitable once you're saying the derivative of f is undefined for x ? 0. the main suitable question you may desire to ask your self is, what's the area of f? this is because of the fact the set of severe numbers could desire to be contained interior the area in accordance to the definition. So the area of f is 0? x < ?. as a result the only severe numbers on your area are: 0, 4/9. maximum suitable.