critical points of .34x3 + 6.95x2 + 46.56x + 101.362
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
first call your function f(x), so you'll have:
\[f(x) = 0.34x^3 + 6.95x^2 + 46.56x + 101.362\]
In order to find the critical points, you'll have to find the first derivative, then you can take 2 conditions to find the critical points which are:
(a) f'(x) = 0
b) f'(x) DNE = Doesn't Exist, and it doesn't exist when you either have x in the denominator, or you have the sqrt of negative number.
So in this case, your function is a polynomial and it's continuous everywhere, so after you find the first derivative, use the first condition to find the zeros of f'(x) , those zeros are your critical numbers, so to find your (y) for the critical numbers, subsitite the (x) that you'll find in the function and you'll get the critical points
Do you understand the concept better now? ^_^