Find the interval on which the following curve is concave upward.?
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
1015745
here goes again .. seems like im gona do the whole hw for u ;) to find the limit of the area that ur gona revolve around the x=-2 axis u have to draw the two equations again and the area enclosed is basicly the shape of a circle (not a perfect circle though) and its located right above the x-axis touching it from below and the center of it is on the y-axis .. now the limit is the intersection of these two euations y=x^2 and y=2-x^2 that is x^2= 2- x^2 ==> 2*x^2 = 2 ==> x^2=1 ==> | x | = 1 ==> x = +1 or x = -1 .. and thats ur interval limits
158550
a curve extending to infinity from zero...
910766
The only interesting point in this curve will be the inflexion point which you find by doing the derivative twice and equating to zero.\n\nf'(x) = 3x^2 + 1\n\nf"(x) = 6x = 0 <===> x = 0\n\nf(0) = -1\n\nThe curve will have an inflexion point at (0,-1).\n\nf(0) < 0, f(1) > 0 then:\n\nThe curve will also have a single real root in [0,1] (x = 0.6823)
152408
Use excel to give delta y/delta x on the curve. Look for a point where this value (the derivative) shoots up in value. That's your point.
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
753539
Since f(x) is a parabola that opens upward, you need the minimum value, which is the y-coordinate of the vertex.  The x-coordinate is -(-12)/(2•3) = 2.  So the minimum y is at 3(2)² - 12(2) + 13 = 1, and your range is y >= 1.\n\nFor the second part, just solve 3x² - 12x + 13 = 28.\n\n3x² - 12x - 15 = 0\nx² - 4x - 5 = 0\n(x-5)(x+1) = 0\nx = 5 or x = -1.
759663
a) The curve you need is called a cycloid. Look that up on the web.\n\nb) The parametric form of the cycloid is\n\nx=12(a-sin a)      y=12(1-cos a) where a is then angular displacement. 24pi (circumference) will be the period and it passes through (0,0)
857047
don't know if you've taken Calc I or not, but if you have, differentiate your equation.  If not, you can look at the next paragraph.You'll get y'=-2x+6.  Remember that deriviatives tell the slope of a function at the x input.  Also remember that quadriatics are U shaped with the co-efficient of the squared term telling if the equation opens up or down.  (positive up, negative down)  Since it's negative, the equation will open down and have a maximum point. (the slope of the max point will be 0) so y'= -2x+6=0 (x=3) will give max value (3,9)\n\nSet your equation equal to zero and solve.\n(-x^2+6x=0)\n-x(x-6)=0\nx=0,6\nsince quadriatics are symetrical around the max/min point, find the point half way between 0 and 6 (x=3) and this will yeild your max/min point. (3,9)
632411
yes u mean like the turning point of a parabolic curve...yea the x-coordinate is equal to -b/2a\nor u can differentiate the equation of the curve and put it equal to zero and solve :\neg  y = ax^2 + bx + c\ndy/dx = 2ax +b \nput this equal to zero and u get x = -b/2a
765813
There's a pretty good online grapher at http://id.mind.net/~zona/ezGraph/ezGraph.html that you can use.  It works basically like a graphing calculator.\n\nIt appears this function should be an upside-down sign curve with an amplitude of 10 and a vertical shift of 9, which means the center of the graph will be at 9, it will go up to 19 and down to -1.  It's not 100% clear what you mean by what you have written after sin.  If you mean the sine of 2pi/15 times x (since the minus zero is irrelevent), you would have a period of 15, so you'd complete 1 complete cycle at 15 radians; the halfway point would be 7.5 radians.
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.
113113
Think of this as an even power function.  Look at the graph of y=-x^2.  Looks like an arch.  So does y=-x^4.  Now you just need to calculate the other parts of this function to match your initial, final, and maximum.  The expression you will be using is y=-(ax+c)^2+b.  The 'a' will change the "roundness" and the 'b' will change the maximum.  The 'c' will change the horizontal position.
735957
Well. You can keep drawing tangents until one which you find is the steepest. From your info, i have the idea that the curve is in the shape of a bell. So perhaps you can try drawing tangents at the mid-range of the curve.\n\nHope this helps=)
922643
It depends on the graph. If the values of each notch differ by x, then the interval is x.\n\nJust look at two numbers on adjacent notches, and subtract.
684833
Convex. \nf" = 3e^x +60x^2 +(1/x)^2 > 0 for all x.
653191
You have to look for the vertical asymptopes. For exapmle in (x^2-5)/((x+2)(x-5)), the vertical asymptopes are the ones that will make the denominator 0, so -2 and 5. With these, as the function approaches them, it will rise to infinity, or unless you have a horizontal or slant asymptope.