Determine the x coordinate of the point where the derivative of y=x2-2x equals 0

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# Determine the x coordinate of the point where the derivative of y=x2-2x equals 0

Derivative function?

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### Determine the x coordinate of the point where the derivative ...

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### Determine the x coordinate of the point where the derivative of y=x2-2x equals 0

y=x2-2x....maebee yu meen x^2-2x???? dy/dx=2x-2, so =0 if x=1

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The line passes through the origin so its equation is y=6x. The equation of the unit circle is x^2+y^2=1, which has centre (0,0) and radius 1. Substituting y=6x in the equation of the circle we have 37x^2=1 and x=+sqrt(1/37). Therefore the y values for the intersections are y=+6sqrt(1/37). The

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### Two adjacent corrals are to be made using 240 ft of fencing. The fence must around the outer perimeter and across the middle. Find the dimensions so that the total enclosed area is as possible.

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### a point is square root of ten units from A(-1,-3). find the point if its coordinate are equal

???????????? "if its koordinut ARE equal...du yu hav 1 or many ?????????????? sound like yu hav a POINT at (-3,-3) & wanna find nuther point with distans twee em 2 points=sqrt(10) sound like a serkel round yer point, so equashun is (x+3)^2 +(y+3)^2=10 me ges yer spozed tu evaluate this

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y=x^3+3x^2-3x-9. The changes of sign determine the minimum number of turning points. There is therefore at least one (sign changes from + to - along the terms of the polynomial). y'=3x^2+6x-3=0 at a turning point. This quadratic can be reduced to x^2+2x=1, and completing the square: x^2+2x+1=2

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### given: y= f(x) = x^3 - 3x +2 determine dy/dx, the second derivative, turning points, min and max, point of inflection

f'(x)=dy/dx=3x^2-3; when 3x^2-3=0 there is a turning-point (the gradient is horizontal). So x^2=1 and x=1 and -1. Second derivative f"(x)=6x. At x=1 f"(1)>0 (minimum) and at x=-1 f"(-1)<0 (maximum). When x=1 y=1-3+2=0; when x=-1 y=-1+3+2=4. The coords for min are (1,0) and ma

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### Determine the first derivative (dy/dx) : y= f(x)=x^3-3x+2

Please note that the title to this question has y=f(x)=x^3-3x+2 but the text of the question has y=f(x)=3x^3-3x+2, which is the equation I have used in my answer. (a) 1. f'(x)=dy/dx=9x^2-3 using the rules of differentiation 2. f"(x)=18x (gives the nature of any turning point) 3 and 4:&nb

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### Maximum volume of a cube

The volume of a cube is L*W*H. So with L = 24-2x, W = 24-2x and H = x you get: V(x)=(24-2x)*(24-2x)*x=((24-2x)^2)*x=(5… The max volume is found when the derivative of V(x) equals zero. V'(x)=576-192x+12x^2=0 Solving that we get two points: x=12 and x=4 We try both solutions in the Volume equa

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### What is the x-value of Y, the point that is three-eighths the distance of segment XZ?

The line segment XZ is the hypotenuse of a right-angled triangle we'll call AXZ, where A is 90 degrees . The point Y is 3/8 along XZ, nearer to X than to Z. By similar triangles the coords of Y will be 3/8 from Y to A and 3/8 from A to Z. AX=AZ=6 (the difference between the x and y coords of points

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### Application of Derivatives =)) Help?

A. Determine where each of the following functions is increasing and where it is decreasing. 1) f(x) = x^2 - 6x + 19 f’(x) = 2x – 6 > 0 for x > 3 Function increases on x > 3 and decreases on x < 3 2) f(x) = 10x - x^2 f’(x) = 10 – 2x > 0 for x < 5 Function

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