anonymous
  • anonymous
hi, need help with finding extremas
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
which part ^^
anonymous
  • anonymous
for example you a a equation y = 3x^2 + x you can find its local extremas first getting its critical points you can get its critical points by deriving it (dx/dy) = 6x + 1 and getting its zeroes 6x + 1 = 0 6x = - 1 x = - 1/6 -1/6 is a critical point now check if its a maxima or minima sustitute a point smaller than -1/6 and greater than - 1/6 to its derivatives.... 6(-1) + 1 -6 + 1 -5 6(0) + 1 1 its a decreasing function before -1/6 because the number smaller than it it gives you a negative number if you substitute it and its a increasing function for the numbers grater than it. we have a clue that it is decreasing then increasing therefore we can imagine a parabole concave upward we can now conclude that the function y = 3x^2 + x have a local minimum at x = -1/6
anonymous
  • anonymous
Thanks How about finding the extrema of y = 10/(x^2 + 1) in the interval (-1, 2).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
We take the derivative of the function and set to zero to isolate those values for x where the function has extrema within the interval. Then\[y'=\frac{d}{dx}\frac{10}{x^2+1}=-\frac{20x}{(x^2+1)^2}\] y' is zero for those values x that satisfy,\[-\frac{20x}{(x^2+1)^2}=0 \rightarrow x=0\]So y has an extremum at x=0. We take the second derivative to determine whether this is a maximum or minimum. doing this, we obtain\[y''=\frac{80x^2}{(x^2+1)^3}-\frac{20}{(x^2+1)^2}\rightarrow y''(0)=-20<0\]so this value for x produces a maximum. That maximum is,\[y(0)=\frac{10}{0^2+1}=10\]
anonymous
  • anonymous
For finding extremas there are several general steps you'e got to take, and if you follow them then you'll be able to solve any extremas probelm: 1) Determine that your function is continuous along in the given interval 2) find the derivative of your function 3) Find the critical points, and you have 2 conditions to find the critical points. (a) take f'(x) = 0 or f'(x) DNE . It's either one of these cases. When you find the zeros of your derivative you've got to check if it's in the interval, if not then it's not considered a critical number. (b) If your condition is f'(x) DNE, then you don't have critical numbers; infact, you will have Vertical asymptotes in this case. 4) After finding the critical numbers, subsitute the numbers in the interval + the critical numbers in the original equation. For example, if my critical numbers qhere 1 and 3, and my interval was (8,0) I'll do the following: - f(1) = .... - f(3) = .... - f(9) = .... - f(0) = .... The largest value/answer that comes out will be considered the Absolute max, and the smallest value/answer that comes out will be considered the Absolute min.
anonymous
  • anonymous
f(8) * instead of f(9) sorry
anonymous
  • anonymous
alot of typos, my bad , I hope you get the point though.

Looking for something else?

Not the answer you are looking for? Search for more explanations.