Respuesta :
Answer:
1+√2, 1-√2
Step-by-step explanation:
Since there are 2 factors with an y in the expression, i will assume that there was a mistake in the question and in fact the first term was y³. With that change, we will have that
g(y) = y³-3y²-3y+9
In order to find the critical numbers of g we need to derivate it and equalize the derivate to 0. We can easily derivate g since it is a polynomial:
g'(y) = 3y² - 6y-3
Since g'(y) is a quadratic function, we can obtain the zeros using the quadratic formula, where a = 3, b = -6 and c = -3:
[tex]r_1 , r_2 = \frac{6 \, ^+_- \, \sqrt{(-6)^2 - 4 * 3 * (-3)} }{2 * 3} = \frac{6 \, ^+_- \, \sqrt{72}}{6} = 1 \, ^+_- \, \sqrt{2}[/tex]
Thus
[tex]r_1 = 1 + \sqrt 2\\r_2 = 1- \sqrt 2[/tex]
Therefore, the critical numbers of g are 1+√2 and 1-√2.
I beleive that the problem just ask for that. If you want the critical values, then we need to evaluate those numbers in g. i will do it just in case
g(1 + √2) = (1+√2)³ - 3(1+√2) - 3(1+√2) + 9 = -1.65685
g(1- √2) = (1-√2)³ - 3(1-√2)² -3(1-√2)+9 = 9.6566
