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anonymous

  • 5 years ago

hi, need help with finding extremas

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  1. anonymous
    • 5 years ago
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    which part ^^

  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    Thanks How about finding the extrema of y = 10/(x^2 + 1) in the interval (-1, 2).

  4. anonymous
    • 5 years ago
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    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,\[y(0)=\frac{10}{0^2+1}=10\]

  5. anonymous
    • 5 years ago
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    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.

  6. anonymous
    • 5 years ago
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    f(8) * instead of f(9) sorry

  7. anonymous
    • 5 years ago
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    alot of typos, my bad , I hope you get the point though.

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