anonymous
  • anonymous
f(x)=-x^2-x-4 Determine a. if f has a maximum value or a minimum b. the value of x at which the max or min occurs c. the max or min value of f
Mathematics
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anonymous
  • anonymous
f(x)=-x^2-x-4 Determine a. if f has a maximum value or a minimum b. the value of x at which the max or min occurs c. the max or min value of f
Mathematics
chestercat
  • chestercat
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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.

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anonymous
  • anonymous
just graph it
myininaya
  • myininaya
the parabola is oven downward so we have an upside down U that means we have a max
myininaya
  • myininaya
open*

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anonymous
  • anonymous
a. maximum b. x = -1/2 c. -9/2
anonymous
  • anonymous
A) derive, set equal to zero. There will be two. Test both critical points. If it goes from positive to negative, it's a max. If it's opposite, it's a min. If there is no change, which there won't be in this instance, it tells you nothing. B) the values you tested in a. C) plug the values in from b and solve for y.
myininaya
  • myininaya
f(x)=-(x^2+x)-4=-(x^2+x+(1/2)^2)-4+(1/2)^2 =-(x+1/2)^2-4+1/4=(x+1/2)^2-15/4 so the vertex=(-1/2,-15/4)
anonymous
  • anonymous
Correction, theres one critical point.

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