Guide :

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

Derivative function?

## Research, Knowledge and Information :

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

Determine the x coordinate of the point where the derivative of y equals x square ... A point whose y-coordinate is 6 lies on ... x2+2x+x0 = 0 Any integer raised to ...

### How calculate the X, Y coordinates of a point. Calculate the ...

How calculate the x,y coordinate of a point. ... The derivative of f(x) at x = x 1 (and y = y 1) ... (x 1) = 0 = 3(x 1) 2 - 12. 3(x 1) 2 = 12

### determine the x-coordinate of the point on the curve y=2x(lnx ...

determine the x-coordinate of the point on the curve y=2x ... 2))? 2. Find the derivative of f(x) = ... two points on the curve where y equals 0 and show that the ...

### Find points on a graph where tangent line is horizontal - YouTube

Feb 04, 2013 · Find the points on the graph of f(x) = 2x^3 ... Find points on a graph where tangent line is horizontal ... Equation of a Tangent Line with Derivatives ...

### X2 - 2x - 5 equals 0? - Find Answers Here! - FinderChem.com

X2 - 2x - 5 equals 0 ... Determine the x coordinate of the point where the ... Determine the x coordinate of the point where the derivative of y=x2-2x equals 0. ...

### What is the x-coordinate of the point of inflection on the ...

... What is the x-coordinate of the point of ... (y''), and the x-values at which y'' equals 0. ... We set our second derivative to 0: ##y''=x^2(2x+6)=0## And we can ...

### CONCAVITY AND INFLECTION POINTS - Georgia Perimeter College

CONCAVITY AND INFLECTION POINTS. Find the Second Derivative of the function, f. ... y-coordinate of the Inflection Point. ... 12x2= 0 . X = 0 . The Second Derivative is

## Suggested Questions And Answer :

### 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

### Find the point(s) where the line through the origin with slope 6 intersects the unit circle.

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 points of intersection are (-sqrt(1/37),-6sqrt(1/37)), (sqrt(1/37),6sqrt(1/37)).  If the unit circle has centre (h,k) the equation is (x-h)^2+(y-k)^2=1 and substituting y=6x we get (x-h)^2+(6x-k)^2=1, which gives the x value of the intersection. So x^2-2xh+h^2+36x^2-12xk+k^2=1; 37x^2-2x(h+6k)+h^2+k^2-1=0. There are only two factors of 37, which is prime, so to factorise rationally we must have (37x+a)(x+b)=0; 37x^2+x(37b+a)+ab=0. Using the quadratic formula: x=(h+6k+sqrt((h+6k)^2-37(h^2+k^2-1))/37. The square root can only be evaluated if the expression is positive, so (h+6k)^2>37(h^2+k^2-1). This requirement applies so that the line intersects the unit circle. When the expression is zero, the line is a tangent to the circle, so there is only one intersection point. h^2+36k^2+12hk>37h^2+37k^2-37; 36h^2-12hk+k^2<37; (6h-k)^2<37 and (6h-k)< +sqrt(37) (=+6.08). This connects the coordinates of the centre of the unit circle: k>6h+sqrt(37). On equality the line y=6x will be tangential to the circle. For example, if h=0 (centre of the circle is on the y axis), k=+sqrt(37) and the unit circle will lie above or below the axis with y=6x running tangentially on the right of the circle; or on the left touching the circle below the x axis. Ideally, we want the square root to be rational so 37-(6h-k)^2=a^2. If a=+1, 6h-k=6 and k=6(h-1); or 6h-k=-6, so k=6(h+1). That gives many possible values for h and k represented by pairs: (1,0), (2,6), (3,12), (0,-6), (-1,-12), (-2,-18), (0,6), (1,12), (-1,0), (-2,-6),  to mention but a few. Using (1,0) in the quadratic: 37x^2-2x(h+6k)+h^2+k^2-1=0 we have 37x^2-2x=0=x(37x-2) giving intersection points (0,0) and (2/37,12/37). The equation for the circle is (x-1)^2+y^2=1 or y^2=2x-x^2. Let's try (2,6). 37x^2-76x+39=0, (37x-39)(x-1)=0 giving intersection points (39/37,234/37) and (1,6). The equation of the circle is (x-2)^2+(y-6)^2=1 or x^2-4x+y^2-12x+39=0. If a=+6, 6h-k=1 or -1, so k=6h-1 or 6h+1. This generates more possible intersection points. There are clearly an infinite number of positions for the unit circle centre (h,k) and an infinite number of intersection points. However, the relationship between h and k so as to produce rational intersection points has been established. k=6(h+1), k=6h+1 are the equations linking the coordinates of the centre of the unit circle. With these equations in mind the quadratic determining the intersection points (x,6x) can be solved: 37x^2-2x(h+6k)+h^2+k^2-1=0. There are four variations of this quadratic because there are four equations linking h and k. Recap There are 2 values of a^2 where a^2=37-(6h-k)^2 and x=(h+6k+a)/37; a^2=1 or 36. When a^2=1, k=6(h+1). The equation of the circle is (x-h)^2+(y-6(h+1))^2=1 and x=(37h+36+1)/37. So the points of intersection are (h+1,6(h+1)), ((37h+35)/37,6(37h+35)/37), ((37h-35)/37,6(37h-35)/37), (h-1,6(h-1)). When a^2=36, k=6h+1. The equation of the circle is (x-h)^2+(y-6h+1)^2=1 and x=(37h+6+6)/37. The points of intersection are ((37h+12)/37,6(37h+12)/37), (h,6h), ((37h-12)/37,6(37h-12)/37). Note that (h,6h) is the result of (37h+6-6)/37 and (37h-6+6)/37. We can check the (h,k) values we used earlier. These were (1,0) and (2,6). We used the formula k=6(h-1) in each case (a=+1), so intersection points for h=1, k=6(h-1)=0, should be x=(h+6k+1)/37, giving (2/37,12/37) and x=(h+6k-1), giving (0,0). For (2,6) h=2 and k=6, giving intersection points x=(2+36+1)/37, giving (39/37,234/37) and x=(2+36-1)/37=1, giving (1,6). The values of h and k are not restricted to integers.

### 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.

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. The picture is a rectangle. The fwnce is across the middle of the rectangle. One side=y. The other side = x+x=2x. How do you know what variable to solve for first and how do you determine the domain? Thanks. You supplied no picture, but it's easy to draw one and solve the problem. Use two horizontal fences and three vertical fences, as shown in the picture I have below. As you can see, the area of one of the smaller corrals is x * y. By substituting the value of y in terms of x, we get x(80 - 4/3 x). Write this as an equation. y = x(80 - 4/3 x) y = 80x - 4/3 x^2 Set this equal to zero to simplify dealing with the left side of the equation. We will be working out the value of x where this graph crosses the x-axis, so y is naturally zero. 0 = 80x - 4/3 x^2 Divide by -4/3. 0 = -60x + x^2 x^2 - 60x = 0 Complete the square. Add (1/2 * -60)^2 to both sides of the equation. x^2 - 60x - 30^2 = 900   <<< that's -30 * -30 Factor the left side. (x - 30)(x - 30) = 900 Take the square root of both sides. x - 30 = ± 30 x = +30 + 30    and  x = - 30 + 30 x = 60          and  x = 0 This tells us that the graph of the equation is a parabola that crosses the x-axis at (0, 0) and (60, 0). The vertex, which is the highest point of the graph, is mid-way between those two points. The value of x at that point produces the maximum area for the smaller corral. Now, we calculate the value of y, using the formula derived in the picture. y = 80 - 4/3 x y = 80 - 4/3 (30) y = 80 - 4 (10) y = 80 - 40 y = 40 The two smaller corrals are 30 x 40; the large corral is 60 x 40 The corresponding areas are 1200 (for each of the two smaller corrals) and 2400 (for the large corral). The length of the fencing is 4 * 30 + 3 * 40 = 120 + 120 = 240. Now that we have the dimensions, we can see that the y sides are longer than the x sides, but the picture served its purpose.

### 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 at sum dont-no point

### Determine the coordinates of the turning points and conclude the maximum turning point of the equation y= (x^2 - 3)(x +3)

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, (x+1)^2=2 so x+1=+sqrt(2), making x=sqrt(2)-1=0.4142 and -(sqrt(2)+1)=-2.4142 the x values of the turning points. The y values are -4(sqrt(2)+1)=-9.657 and 4(sqrt(2)-1)=1.657. So the turning points are (0.4142,-9.657) and (-2.4142,1.657). Further differentiation tells us what type of turning points we have: y"=6x+6. When x=0.4142 this is positive so the point is minimum; when x=-2.4142 it's negative, so maximum. Therefore, the maximum turning point is (-2.4142,1.657) or (-(sqrt(2)+1),4(sqrt(2)-1)).

### 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 max (-1,4).

### 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: f'=0 at a turning point, so 9x^2=3 and x^2=1/3 so x=±1/√3 or ±√3/3 in rational form. When x>0 f">0 and when x<0 f"<0. This means that x=√3/3 is a minimum and x=-√/3 is a maximum.  5. There is no point of inflexion. 6. The graph is shown below. The intercepts are y=2 and the root(s) of 3x^3-3x+2=0. They only point where the curve intercepts the x axis is about x=-1.24. If f(x)=x^3-3x+2 was the intended equation, the graph will look similar (not the same) but all the answers will not fit. Nevertheless I think you can work out the answers for yourself given the general idea of how to do so in this solution. The major difference is that x^3-3x+2 factorises: (x+2)(x-1)^2 with roots x=-2 and x=1 (x intercepts, so the curve cuts or touches the x axis twice). Also the max and min are at x=±1 (max at x=-1 and min at x=1). The graph below of x^3-3x+2 clearly shows the intercepts with the curve just touching (not intercepting) the x axis at x=1, while x=-1 is a clear intercept. The turning points are (-1,4) (max) and (1,0) (min). If you would like a full solution to this, add a comment to this answer and I will answer using a comment reply.

### 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 equation: V(4)=1024 V(12)=0 So the correct answer is 1024 when x=4.

### 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 X and Z). 3/8 of 6=9/4=2.25. Take 2.25 from the y value of X: 5-2.25=2.75. The x value for Y is x value of X plus 2.25=1+2.25=3.25. Y is (3.25, 2.75). Another way to look at this is to derive the equation of the line passing through X, Y and Z. The slope is -1, because it's derived from (-1-5)/(7-1)=-6/6=-1. The intercept is found by entering the coords of X or Z into y=-x+b (b is the intercept). Put in X(1,5) and b=6, so y=6-x. The point Y fits on this line. The x value is 3.25.

### 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 increases on x < 5 and decreases on x > 5 B. Determine the critical values of each of the following functions: 1) f(x) = x^2 - 16x f’(x) = 2x - 16 = 0 at x = 8 Critical value is f(8) = 8^2 - 16*8 = 64 – 128 = -64 Crit value: -64 2) f(x) = x^3 – 2 f’(x) = 3x^2 = 0 at x = 0 Critical value is f(0) = 0 - 2 = -2 Crit value: -2 C. Find all relative extreme points of each of the following functions: 1) f(x) = x^2 - 20x f’(x) = 2x – 20 = 0 at x = 10 f(10) = 10^2 – 20*10 = 100 – 200 = -100 Minimum point is at (10, -100) 2) f(x) = x^3 - 3x – 2 f’(x) = 3x^2 – 3 = 0 at x = 1 and at x = -1 f(1) = 1 – 3 – 2 = -4, f(-1) = -1 + 3 – 2 = 0 Minimum point is at (1, -4)       Maximum point is at (-1, 0) 3) f(x) = -x^3 - 3x^2 + 7 f’(x) = -3x^2 – 6x = 0 at x = 0 and at x = -2 f(0) = 0 – 0 + 7 = 7, f(-2) = -(-8) – 3(4) + 7 = 3 Minimum point is at (-2, 3)       Maximum point is at (0, 7) 4) f(x) = x^4 - 2x^2 + 3 f’(x) = 4x^3 – 4x = 0 at x = 0 and at x = 1 and at x = -1 f(0) = 0 – 0 + 3 = 3, f(1) = (1) – 2(1) + 3 = 2,  f(-1) = (1) – 2(1) + 3 = 2 Minimum points are at (-1, 2) and (1, 2)       Maximum point is at (0, 3)