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# find all positive values of x that satisfy x * the x root of x^2=x^x

x*the x root of x^2=x^x

## Research, Knowledge and Information :

### (x − 1)/(x + 1) + (x)/(x − 2) < 1 +10/((x − 2)(x - Socratic

... the quotient is negative On (3, oo), the quotient is positive. We want the values of x that ... SOCRATIC ... you find the values of x that satisfy ...
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### find the range of possible values of x that satisfy the ...

... [0,2]. Find all values of c that satisfy the ... Find all positive values of x that satisfy x multiplied by the x root of x^2 = x^x trig find all the values of ...
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### find all the values of x that satisfy the equation: 2sin^2(x ...

find all the values of x that satisfy the equation: ... (5x/2sqrtx+2) ... calculus Find all positive values of x that satisfy x multiplied by the x root of x^2 = x^x
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### Find the range of values of x which satisfy the inequality x ...

Find the range of values of x which satisfy the inequality x^2-x ... one factor must be positive and ... ≤0 True if x is between the roots, or a root, so ...

### Solving inequalities - Mathematics resources - www.mathcentre ...

Solving inequalities ... of values that an unknown x can take and still satisfy ... 2. If we write |x| < 2 we mean all points a distance less than 2 units from O.
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### Quadratic Equations and the Discriminants

Quadratic Equations and the Discriminants ... 5 What are all real values of x which satisfy 2 ... root is m less than the second root of the original equation. Find b ...
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### Find all values of x in the interval [0,2pi] that... - OpenStudy

Find all values of x in the interval ... Find all values of x in the interval [0,2pi] that satisfy the ... But after you divide by 2 you would square root both ...
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### What values of $x$ satisfy $(\sin x ... - Quora What values of [math]x$ satisfy [math](\sin ... one possible positive real root. [math ... they are only defined for positive values of the base and the value ...
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### Find all values of x in the interval [0,2π ] that satisfy cos ...

Find all values of x in the interval [0,2π ] that satisfy cos^(2)x+sinx=1. ... Positive: 37 %. Answer #2 | 19/08 2014 09:49

## Suggested Questions And Answer :

### show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)] Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)). First find where sqrt(x^2 + 1) is continous on. We know that for square roots, the number has to be greater than or equal to zero (definitely no negative numbers). So set the inside greater than or equal to zero and solve for x. You'll get an imaginary number because when you move 1 to the other side, it'll be negative. So, this means that the number inside the square root will always be positive, which makes sense because the x is squared and you're adding 1 to it, not subtracting. There would be no way to get a negative number under the square root in this situation. Therefore, since f(x) is continuos everywhere, (-infinity, infinity), then f(x) is continuous on [0, sqrt(8)]. Now you have to check if it is differentiable on that interval. To check this, you basically do the same but with the derivative of the function. f'(x) = (1/2)(x^2+1)^(-1/2)x2x which equals to f'(x) = x/sqrt(x^2+1). So for the derivative of f, you have a square root on the bottom, but notice that the denominator is exactly the same as the original function. Since we can't have the denominator equal to zero, we set the denominator equal to zero and solve to find the value of x that will make it equal to zero. However, just like in the first one, it will never reach zero because of the x^2 and +1. Now you know that f'(x) is continous everywhere so f(x) is differentiable everywhere. Therefore, since f(x) is differentiable everywhere (-infinity, infinity), then it is differentiable on (0, sqrt(8)). So the function satisfies the two hypotheses of the Mean Value Theorem. You definitely wouldn't have to write this long for a test or homework; its probably one or two lines of explanation at most. But I hope this is understandable enough to apply to other similar questions!

### complex, rational and real roots

The quintic function should have 5 roots.  The changes of sign (through Descartes) tell us the maximum number of positive roots. Since there are two changes of sign there is a maximum of 2 positive roots. To find the number of negative roots we negate the terms with odd powers and check for sign changes: that means that there is at most one negative root, because there's only one sign change between the first and second terms. Complex roots always come in matching or complementary pairs, so that means in this case 2 or 4.  Put x=-1: function is positive; for x=-2 function is negative, so there is a real root between -1 and -2, because the x axis must be intercepted between -1 and -2. This fulfils the maximum for negative roots. That leaves 4 more roots. They could be all complex; there could be two complex and two positive roots. Therefore there are at least two complex roots. To go deeper we can look at calculus and a graph of the function: The gradient of the function is 10x^4-5. When this is zero there is a turning point: x^4=1/2. If we differentiate again we get 40x^3. When x is negative this value is negative so the turning point is a maximum when we take the negative fourth root of 1/2; at the positive fourth root of 1/2 the turning point is a minimum, and these are irrational numbers. The value of the function at these turning points is positive. The fourth root of 1/2 is about 0.84. and once the graph has crossed the x axis between -1 and -2, it stays positive, so all other roots must be complex. The function is 12 when x=0, so we can now see the behaviour: from negative values of x, the function intersects the x axis between -1 and -2 (real root); at x=4th root of 1/2 (-0.84) it reaches a local maximum, intercepts the vertical axis at 12 until it reaches 4th root of 1/2 (0.84) and a local minimum, after which it ascends rapidly at a steep gradient.  [Incidentally, one way to find the real root is to rearrange the equation: x^5=2.5x-6=-6(1-5x/12); x=-(6(1-5x/12))^(1/5)=-6^(1/5)(1-5x/12)^(1/5) We can now use an iterative process to find x. We start with x=0, so x0, the first approximation of the solution, is the fifth root of 6 negated=-1.430969 approx. To find the next solution x1, we put x=x0 and repeat the process to get -1.571266. We keep repeating the process until we get the accuracy we need, or the calculator reaches a fixed value. After just a few iterations, my calculator gave me -1.583590704.]

### find the first 50 positive roots of cot(x)-.45x for f(x)=0

I do not have have access to Mathcad so I'm unable to generate 50 roots, so I have provided the first seven roots only. (Method: trial and error.) Plotting the graph of cot(x) and 0.45x on the same axes helps to find coarse solutions. Where the line cuts the curve is a solution or root of cot(x)-0.45x. The cot curve is cyclic and cuts the x axis at odd multiples of 90 degrees or (pi)/2 radians. So there is always a solution between 180n and 180n+90 degrees, n(pi) and n(pi)+(pi)/2 radians, where n>0. If x is in degrees, these are the first 7 positive solutions for x: 11.2114, 180.7046, 360.3533, 540.2357, 720.1768, 900.14145, 1080.1179. You can see that as the root gets larger its value gets closer to a multiple of 180 degrees. If x is measured in radians, these are the first 7 positive solutions for x: 1.1082, 3.6843, 6.6076, 9.6511, 12.7391, 15.8473, 18.9662 [As the root gets larger the difference between itself and the previous solution gets closer to (pi) (3.14 approx). 22/7 is slightly bigger than pi, but every 7th root in the succession of roots is going to be roughly 22 larger than the 7th previous root. This fact can help us to quickly find every 7th root: 14th, 21st, 28th, 35th, 42nd, 49th. The 14th root should be roughly 19+22=41. Let's see what it actually is: 40.895. Close. Add 22 to this to get the 21st root: 62.9. The actual value is 62.8672. To get the 49th root we need to add approximately 28(pi) or 88=150.9. Actual: 150.8112. Something similar happens when x is in degrees.]

### I just need help getting started on this problem. I don't even know how to begin!

The graph is centered over the line x=-3. This line acts like a mirror so that the two halves of the inverted U-shaped graph are reflected in it. The graph has to be shifted 3 units to the right so that the line x=-3 becomes the line x=0, which is the y (f(x)) axis. The equation of the graph changes to f(x)=-0.5x^2+8. When x=0 f(0)=8, and this value on the vertical axis is the highest point of the curve (vertex). To picture f(x+k) you need a picture, a moving picture. Think of the horizontal axis, the x axis, and the line f(x)=8, which is a line parallel to the x axis a distance of 8 units above it. The two lines resemble a track, like a rail track. The curve  is, as we've established, an inverted U shape where the arms of the U are moving further apart the further away the curve is from its vertex. The curve has a constant shape and remains in contact with the line f(x)=8. We start with the curve right in the middle so that the vertical axis bisects the U curve. This is when k=0.  Now, we're going to slide the curve so that its vertex runs along the track f(x)=8 left and right. The curve cuts the axis at two points separated by 8 units, the gap between the zeroes of the function. When the curve is centrally positioned, the points on the x axis are -4 and 4 and k=0. Slide the curve to the left (negative side) and this is equivalent to positive values of k; slide it to the right and we're into negative values of k. If we move the y (f(x)) axis with the curve it becomes the movable axis of symmetry, the mirror I mentioned earlier. The points where the curve cuts the x axis remain 8 units apart. When the curve moves leftward one unit, the value of k increase by 1, and rightward one unit when k decreases by 1. The vertical line x=2 is unmovable, but as the curve slides from left to right this line touches or cuts through the curve. Consider only the part of the curve lying on or above the x axis. When does it touch the vertical line? Move the curve so that the right part of the curve just touches the line. This is the zero with the higher value. So it must be when x=2 is a root, i.e., -0.5(2+k)^2+8=0. 0.5(2+k)^2=8, so (2+k)^2=16, 2+k=+4, and k=4-2 or -4-2, which is 2 or -6. If we slide the curve past the line x=2, till the left part of the curve touches the line on the right side, the point where it touches is the other root, 8 lower than the the right-hand root. So if 2 was the right root then 2-8=-6 is the left root, and if 2 was the left root then 10 is the right root. So we have the functions f(x)=-0.5(x+2)^2+8 and f(x)=-0.5(x-6)^2+8, where the values of k have been substituted. When the curve is on the left of x=2, the roots of f(x+2) are -6 and 2 and when on the right the roots of f(x-6) are 2 and 10. In between these values of k, 2 and -6, the curve touches or is above the x axis, so this the range for k: -6 Read More: ...

### how do I solve for x with this equation : 2xcubed + 13xsquared - 24x - 16 = 0?

2x^3+13x^2-24x-16=0 has no rational roots, and so doesn't factorise. So how can we possibly find the roots? One way is to use systematic trial and error. You need a calculator for this. It can take a little time to get all three roots (a cubic equation has a maximum of three). You can also draw a graph of the function y=2x^3+13x^2-24x-16 sufficiently accurately to see where the graph cuts (intercepts) the x axis. These are the roots. If you do this you may be able to see that the smallest root is between -8 and -7, another between -1 and 0, and another between 1 and 2. Starting at x=-8, -1 or 1 calculate y and note the result. Then change the value by a small increment (e.g.,0.2) so you have x=-7.8, -0.8 or 1.2, and recalculate y. Note the result and particularly note where there is a sign change from positive to negative or vice versa. A sign change indicates that the root is between two values. For example, putting x=-1, y=19 and putting x=0, y=-16. So the sign change is positive to negative for the nearest root. You can narrow the search once you know between what values a sign change occurs and then you can change the increment to the next decimal place. When I did this exercise for each estimated root I was able to find them to a higher accuracy: x=-7.8921, -0.5280 and 1.9200.

### what are the roots of the polynomial equation 2x^3+2x^2-19x+20=0

With questions of this sort, find by trial a value of x for which the equation has a positive result and another for which it has a negative result. You then know that a root must lie in between those x values. If you find a value making the result zero, that's even better, because you found a root straight away. Because this is a cubic equation you only need to find one root, because you can divide by that root to get a quadratic, which is easier to solve. As it happens by trial I find that x=-4 is a root, so I can use synthetic division or algebraic long division to divide the equation by x+4 or the root -4. I get the quadratic to be: 2x^2-6x+5=0. The quadratic has no real roots, because the quadratic formula gives a negative for which there's no square root, so x=-4 is the only real root. However, the quadratic formula gives us complex roots: (6+/-2i)/4=(3+/-i)/2. These complex roots are: x=(3+i)/2 and (3-i)/2, and the real root is -4.

### find all positive values of x that satisfy x * the x root of x^2=x^x

x^2=x^x...anser...x=2, x=1 maebee x=0, depend on how yu define 0^0

### If M=√2÷√(×+3)(×-2) determine the value of × for which M is real

This is easier than it looks. The square root term including x is only meaningful in this context when the quantity under the square root is strictly positive, so (x+3)(x-2)>0.  So we have just 2 conditions when this is positive: x+3 and x-2 are both positive; and x+3 and x-2 are both negative. Both positive means x+3>0 and x-2>0, so x>-3 and x>2. Since 2>-3, x>2 satisfies both inequalities.  Now when are they both negative? x+3<0 and x-2<0, so x<-3 and x<2. Since -3<2 then x<-3 satisfies both inequalities. So there you have it: M is real when x>2 or x<-3. But if √2x is the numerator rather than √2 and the expression is √(2x) rather than (√2)x, we must eliminate x<-3 leaving x>2 as the solution.

### In your own words, explanation what it means to find the absolute value of each number, use a number line example ti support your explanation

The absolute value of any number is its magnitude regardless of whether it is positive or negative. So absolute value is always positive. It's the same as taking the positive square root of the number squared. So the absolute value of 10 and -10 is 10, because both have a square of 100. If we take a straight line and mark its centre point as 0, then numbers to the left of centre are negative and those to right of centre are positive. Taking the absolute value is the same as measuring how far the number is from the centre point. -10 is 10 away from zero and so is +10.

### find the asymptotes of the hyperbola

If we work out y in terms of x we get y=2+/-15*sqrt(((x+2)/20)^2-1), which only exists when (x+2)^2=>400, i.e., x=>18 or x<=-22. The asymptotes must therefore, it seems, be associated with these values. At x=18 or -22, y=2. The line y=2 represents a definable limit of the function that is attainable at points (18,2) and (-22,2), whereas an asymptote is not attainable. If we write x in terms of y we get x=-2+/-20*sqrt(1+((y-2)/15)^2), we can see that the square root term can never be negative. As y gets bigger, however, the square root approaches (y-2)/15, so x approaches -2+/-20(y-2)/15 which is -2+/-4/3(y-2). Similarly, y approaches 2+/-15(x+2)/20, which is 2+/-3/4(x+2). That is, both x and y tend to infinity positively and negatively, and an asymptote cannot be defined, unless we take the positive and negative slopes, defined by +/-3/4 as x and y tend to infinity. The asymptotes then form an X where the centre of the X is the point (-2,2), because x=-2 is halfway between x=18 and -22. The asymptotes are therefore two straight lines with opposite slopes in the form y=ax+b. Both lines intersect at (-2,2) so we can substitute these values for x and y to find b in each case. When we do this we get b=7/2 and 1/2, giving us the equations: y=(3/4)x+7/2 and y=-(3/4)x+1/2 or 4y=3x+14 and 4y=2-3x. [The shape of the curve is interesting. The line y=2 is a reflector, on the positive side a curve emerges from (18,2) and moves off to the right as x and y get larger and larger. Underneath the line is its reflection which cuts the x axis at about 18.177 and continues to the right reflecting the upper part of the curve. The two halves together form U shapes, lying on their sides, with outward curving arms, and with the centre of the U's on the line y=2. The y axis is also a reflector, and the two halves of the curve are reflected on the left with the same hyperbolic shape as on the right but in the opposite direction. The x axis is cut at x=-22.177.]