Test for concavity and inflection points?
917389
Concavity is found from the second derivative.\nf'(x) = 6x^2 + 6\nf''(x) = 12x\nSet this equal to zero\n12x = 0 means x = 0\nSince 12x < 0 when x < 0 and 12x > 0 when x>0 that means x = 0 is an inflection point.\n\nTo find the actual point, you will need to replace x with 0 in f(x) and simplify.
818186
.  The y intercept is simply where the curve intersects the y axis, which means x=0.  Just plug in zero for x and find what the function is equal to.\n.  The local maxima and minima occur where the first derivative (the slope of the curve) is equal to zero.\n.  Points of inflection are where the rate of change in slope (the second derivative) goes from positive to negative or vice versa, so take the second derivative.\n.  The value of the function is increasing with x where the slope, (the first derivative) is positive.  The function is decreasing where the slope and first derivative are negative.\n.  The curve is concave up when the slope is increasing with x, and concave down where the slope is decreasing; so define the intervals over which the second derivative is positive or negative, and you will have define the concave up and concave down regions.
736099
plot the coordinates then do the vertical line test.  if the vertical line intersects more than two points on the graph then it is not a function.  each set of coordinates you have listed are functions (i free hand it with no graphing paper or calc) so forgive me if im wrong.  also refer to www.coolmath.com. it is a good site for breaking down math
807339
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1\nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n.
959832
Pick a point in the shaded region, plug that coordinate into the equations. If the result you produce is true, then you have done your job well!
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
320591
The first derivative of this is\n\n10x^4 - 32x^3 + 21x^2 - 20x + 10\n\nSet that equal to zero and solve for the FOUR solutions. That will give you the x values of the four turning points that are the maxima and minima of the original equation. \n\nThe second derivative will serve two purposes. It will enable to you to find the thtree inflection points, and also enable you to tell which of the four turning points of the curve are maxima and which of them are minima.\n\nThe second derivative is \n\n40x^3 - 96x^2 + 42x - 20\n\nSet that equal to zero and solve. The three solutions give you the x-values of the three inflection points. To use the second derivative to find which turning points are maxima and which are minima, plug in the value of x for each turning point into the second derivative. If the value of the second derivative is positive, the turning point in a minimum. If it is negative, the turning point is a maximum.
463511
Use the double angle identities.
951330
Try this site. Inequalities and absolute value are really easy once you get it. you may also try posting some examples, and we can help you. Good luck on your next test!\n\nhttp://www.purplemath.com/modules/ineqsolv.htm\n\nhttp://www.purplemath.com/modules/absolute.htm
809962
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1 \nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n\n2. f(x)=(x^2 + x)^2/3, [-2,3]\n\n3. f(x)= [6-4x], [-3,3]\n\n/
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
1112728
to draw lines u hav to select 2 points at a time (as no 3 r collinear)\n\nthus no of selections of such points is nC2= n!/ [ 2! (n-2)!]
729410
You can  find two different values for x that are easy to find 1/7 of, such as 7 and 14; see what y would be at those points [ -4, -3]; then plot the points in either order and connect the dots.  It is smart to also plot a third point to reduce the chance of error, because if one of the three is wrong, they won't line up.
1122353
Check out PhotoImpression 4
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
424777
I'd think you could find two points with the same single variable like temperature, or something easier to visualize like height, but two variables is questionable. Suppose you had 2 altimeters that you could position remotely (and they automatically stayed at opposite points). Idealize height as being a smooth function of location. \nInitially the altimeters indicate unequal heights. When you move them one way the difference increases. Then move the opposite way to decrease the difference. Move a little bit, then reassess the relative changes, find a new direction if needed, and make another small step toward decreasing difference. Now mentally superpose the two surfaces you are surveying. If there are no height maxima or minima and the surfaces intersect, repeating this procedure should find a pair of zero-difference points. Now you can follow the contour of intersection of the surfaces and the difference remains zero. If the surfaces diverge in every direction there must be an altitude maximum or minimum in one or both survey areas; move until the extreme is off the survey area and there is overlap, and repeat the zero-finding procedure. There is an overwhelming probability that such a pair of areas, and therefore of points, can be found. Earth does not have a situation where one hemisphere's altitude range is completely separate from the other hemisphere's. The same goes for temperature and pressure. \nWhat about a second variable, say pressure? Suppose it too can be mapped to two intersecting surfaces. Project the zero-height-difference and zero-pressure-difference contours onto the same plane. If the two contours intersect anywhere, the two-variables-equal point pair has been found. If they don't, you can't find one here; move to a whole new area.\nBy this logic you're extremely likely to find a pair of one-variable equality points, and pretty likely to find a two-variable pair of points. I know this is not a good mathematical answer, which should be a proof or disproof, but then it may not be provable or disprovable.
844185
square-ish in shape\ncome in different sizes\nmany varieties