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

Find the first 50 positive roots of the given function usig mathcad

## Research, Knowledge and Information :

### Math Questions With Answers (2) - analyzemath.com

Math Questions With Answers (2) ... = 0 Roots: x = 1 and x = 4 Check answer: ... We first note that the zeros of f are x = 0 and x = 6.

### Project 2 Part #1: Horner’s Rule and Root Finding

... Horner’s Rule and Root Finding ... f x x x( ) cot( ) 1.275 Here, the roots of this equation are related to the ... You are to find the first 50 positive roots ...

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... if cos x = 2/9 and tan x < 0, without using a calculator? ... tan and cot are positive in 1 and 3. ... What is sin x if tan (45°+x) = 2 ...

### Trigonometric functions - Wikipedia

... definition of the trigonometric functions for angles between 0 ... of the trigonometric functions to all positive and ... (0°), sin(30°), sin(45 ...

### Finding Roots - Free Math Help

... Finding Roots. ... the roots are the values of x for which \(f(x)=0 ... (x = 2\), because that value produces \(f(x)=0\). Of course, it's easy to find the roots ...

### How to Graph a Cotangent Function - dummies

How to Graph a Cotangent Function. ... You can find the parent graph of the cotangent function f(x) = cot x, ... The first one comes at 0 and then repeats every pi ...

### Calculation of The Trig - Pellissippi State Community College

Calculation of The Trig. ... the tedious task of finding the trig ratios of angles other than 0, 30°, 45°, 60°, ... and find angles: x, ...

### Math 113 HW #9 Solutions

Math 113 HW #9 Solutions §4.1 50. Find the absolute maximum and absolute minimum values of f(x) = x3 −6x2 +9x+2 on the interval [−1,4]. Answer: First, we ﬁnd ...

## Suggested Questions And Answer :

### 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.]

### 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 two square roots?

All numbers have two square roots. The principle positive square root as well as the negative square root. The reason for this is because when multiplying a negative number by another negative number we get a positive result. So for example: 8*8 = 64 but. . . -8*-8 also = 64 So the square root of 64 is both positive 8 and negative 8 The first few should be obvious: √64 = ± 8 √169 = ± 13 √256 = ± 16 √400 = ± 20 √1600 = ± 40 For the next you need to break apart the decimals and think of them as fractions: √0.04 = √(4/100) = √4 / √100 = ± 2/10 = ± 0.2 √0.36 = √36/100 = ± 6/10 = ± 0.6 √1.21 = √121/100 = ± 11/10 = ± 1.1

### Guyssss please help me "Create a problem situation in which only the positive solution is meaningful" what am I going to do? I can't understand it

First, you need to understand the question, right? You need to know what gives you more than one solution. The answer is an equation which contains a polynomial of at least degree 2. That means a quadratic equation or a cubic equation, or more, that has more than one solution. A typical example is a quadratic that has a positive root and a negative root: (x-a)(x+b)=0 or x^2+x(b-a)-ab=0, where a and -b are the roots. Next we need a situation which implies that we need a positive solution only. An example is a geometrical figure like a triangle or rectangle which must, of course, have sides of positive length. Now we have to think of a situation which introduces the square of a number. In the case of geometrical figures this is likely to be the area. Now we have to invent a problem. We start with a solution. Here's an example: A rectangle has dimensions 9 by 4 so its area is 36 (never mind the dimensions at the moment). Then we create an unknown, x. Now we use x to define the dimensions of the rectangle. If we say that the length is 5 more than the width and the area is 36, then the solution is: let x=width then length=x+5, then the area is x(x+5)=36. So x^2+5x=36, or the quadratic equation: x^2+5x-36=0=(x-4)(x+9). The roots are 4 and -9, but we can reject -9. Now we have the basics of the problem, let's make it more interesting. First add dimensions. Let's use feet. Replace the rectangle with a familiar rectangular object: a box, swimming pool, a garden, etc. OK here's the question. A swimming pool's length is 5 feet more than its width. What is its perimeter if the area of the base is 36 square feet? We know how to find the width (4 feet) from the solution above, so we find the length (4+5=9 feet). The perimeter adds an extra to the problem, because it uses the length and the width. Perimeter=2(length+width)=26 feet.

### 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!

### integral from 0 to infinity of (cos x * cos(x^2)) dx

The behaviour of this function f(x)=cos(x)cos(x^2) is interesting. The integral is the area between the curve and the x axis. If the functions cos(x) and -cos(x) are plotted on the same graph, the latter form an envelope for f(x). Between x=(pi)/2 and 3(pi)/2, the curve has 4 maxima and 3 minima; between x=3(pi)/2 and 5(pi)/2 there are 7 maxima and 6 minima; between x=(2n-1)(pi)/2 and (2n+1)(pi)/2 there are 3n+1 maxima and 3n minima (integer n>0), a total of 6n+1. (These are based purely on observation, and need to be supported by sound mathematical deduction.) As n becomes large the envelope appears to fill as the extrema become closer together. As x tends to infinity n also tends to infinity. The envelope apparently has as much area above (positive) as below (negative) the x axis so the total area will be zero as the positive and negative areas cancel out. The question is: do the areas cancel out exactly? As x gets larger, the curve starts to develop irregularities and patterns, but it stays within the envelope, and positive irregularities appear to be balanced by negative irregularities, so the overall symmetry appears to be preserved. f(x)=0 when cos(x)=0 or cos(x^2)=0, which means that x=(2n-1)(pi)/2 or sqrt((2n-1)(pi)/2), where n>0. Between 3(pi)/2 and 5(pi)/2, for example, we have sqrt(3(pi)/2), sqrt(5(pi)/2), ..., sqrt(13(pi)/2), because sqrt(13(pi)/2)<3(pi)/20, and this lies between (2n-1) and (2n+1); so n is defined by 2n-1<(2m+1)^2(pi)/2<2n+1.  For m-1 we have 2(n-z)-1<(2m-1)^2(pi)/2<2(n-z)+1, where z is related to the number of zeroes in the current "batch". For example, take m=3: 2n-1<49(pi)/2<2n+1; 49(pi)/2=76.97 approx., so 2n-1=75, and n=38. Also 2(n-z)-1<25(pi)/2<2(n-z)+1 so, because 25(pi)/2=39.27 approx., 2(n-z)+1=41, n-z=20, and z=18. When m=2, 2n-1=39, n=20; 2(20-z)-1<9(pi)/2<2(20-z)+1; 2(20-z)-1=13, 20-z=7, z=13. The actual number of zeroes, Z, including the end points is 2 more than this: Z=z+2. Now we have an exact way to calculate the number of zeroes in each batch. So Z and n are both related to m. The number of extrema, E=Z-1=z+1. In fact, E=int(2(pi)(m-1)+1), where int(a) means the integer part of a, so as m increases, there are proportionately more extrema over the range (2m-1)(pi)/2 to (2m+1)(pi)/2. The figure of 6n+1 deduced earlier by observation approximates to the mathematical findings, because 2(pi) is approximately equal to 6. But we still need to show, or disprove, that the areas above and below the x axis are equal and therefore cancel out. Unfortunately, if we consider the area under the first maximum (between x=(pi)/2 and sqrt(3(pi)/2)), and the area above the first minimum (between x=sqrt(3(pi)/2) and sqrt(5(pi)/2)), they are not the same, so do not cancel out. More...

### Find all of the zeros: x^5 – 3x^4 – 15x^3 + 45x^2 – 16x + 48 = 0

Notice a strange thing about the polynomial? The ratio of the coefficients of the first two terms, the middle two and the last two are all the same: 1:-3. The second term's coefficient is -3 times the first's; the fourth term's coefficient is -3 times the third's; and the last term's coefficient is -3 times the penultimate's. Let's start factorising! x^4(x-3)-15x^2(x-3)-16(x-3)=0; so (x^4-15x^2-16)(x-3)=0 and we can take it further: (x^2-16)(x^2+1)(x-3)=0, and further still: (x-4)(x+4)(x-i)(x+i)(x-3)=0. So the roots are -4, 3, 4, i, -i. Two imaginary roots, just like you thought!

### factor using the trial-and-error method: 12x squared - 5x -3

We need to be careful with this function, because it's quadratic and therefore it could have two roots, two solutions for x when the function=0. These solutions determine the factors. If we plot the graph of the function roughly we can see that when x is large and positive or negative, the function is positive. If we put x=0 we can see where the graph crosses the vertical axis, -3 in this case. To get to -3 the graph must intersect the x-axis at two points. The graph has a U shape. Let's put a couple of values into the function. At x=1 the function is 4. At x=-1 it's 14. So we know it cuts the x-axis twice between x=1 and -1. That gives us a start. Using a calculator to speed things up, we can try values in these limits. Let's go for x=0.5 and -0.5. The two values are respectively are -2.5 and 2.5. What does that tell us? Between x=0 and 0.5 the function goes from -3 to -2.5, and between -0.5 and 0 it goes from -3 to 2.5. So in one case the graph doesn't cross the x-axis but in the other case it does because we've gone from negative to positive between x=0 and -0.5. We need to look first for the solution between -0.5 and 0, so we can try values between -0.4 and -0.1. When we do this we find we go from positive to negative between -0.4 and -0.3. So using trial and error we continue to the next decimal place and try values between -0.39 and -0.31. This time the sign change takes place between -0.34 and -0.33. This is beginning to look like -1/3. Let's put this value into the function. Yes! x=-1/3 is a value that makes the function zero. Therefore one factor is (3x+1). The other factor can be found by inspection. It must start with 4x to make 12x^2 in the function, and the numerical part must be -3 because we have -3 in the original function as the last term and -3*1=-3. So factorisation gives us (3x+1)(4x-3).

### first derivative test

After you find the first derivative, set it equal to zero that will give you the potential candiates for absolute extrema. Then construct a sign chart of the first derivative for those numbers where you pick a number less than or between those numbers and then plug them into for x in the first derivative. Solve the equation and see if their sign is positive or negative then. If it is negative then positive it is min. If it is positive then negative it is min. Now that is local extrema to find absolute extrema take the values for the local extrema and plug those into the ORIGINAL EQUATION and find out which one is the biggest and smallest values.

### What is the answer in 5/7a-3 + 3/a < 5a?

(5/7a)-3+3/a<5a. Multiply through by 7a: 5-21a+21<35a^2 assuming a>0. 35a^2+21a-26>0 has irrational roots one of which is negative, which reverses the inequality. The positive root is 0.61261 approx, and the negative root is -1.21261. To satisfy the inequality, a<-1.21261 or a>0.61261. To appreciate this better, 35a^2+21a-26 is a parabola which cuts the a axis (horizontal) at the roots. It's U-shaped so the vertex is below the intersection points. This means the parabola equation is negative between the roots, so it's positive outside the roots. Therefore apositive root satisfies the inequality. It's possible the question is supposed to read differently. Parentheses help to make the intention clear. If the first term is 5/(7a-3) the result is a cubic which doesn't factorise and the solution is more complicated involving an irrational root. Another interpretation: 5/(7a-3+(3/a))<5a; 5a/(7a^2-3a+3)<5a; 1/(7a^2-3a+3)<1; 1<7a^2-3a+3; 7a^2-3a+2>0. This has no real roots and is always positive for all a. So this interpretation is going nowhere!

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