Guide :

find the zeros of each function state the multiplicity of multiple zeros

y = x(x-8)^2

Research, Knowledge and Information :


Polynomial Graphs: Zeros and their Multiplicity ... - Purplemath


The zeroes of the function (and, yes, "zeroes" is the correct way to ... The odd-multiplicity zeroes might ... But if I add up the minimum multiplicity of each, ...
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find zeros and state multiplicity - YouTube


Feb 21, 2009 · find zeros and state multiplicity ... Finding the Zeros and Multiplicity of Each Zero ... How to find the zeros of a polynomial function by factoring ...

Find the zeroes of each function. State the multiplicity of ...


Find the zeroes of each function. State the multiplicity of multiple zeroes. y=(x+3) ... Find the zeroes of each function. State the multiplicity of multiple zeroes ...
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Find the zeros of the polynomial function and sta... - OpenStudy


Find the zeros of the polynomial function and state the ... function and state the multiplicity of each ... pattern of zeroes, any multiple of g(x ...
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Find the zeros of the polynomial function and state the ...


Find the zeros of the polynomial function and state the multiplicity of each ... Find the zeros of the polynomial function and state the ... Write a function ...
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Zeros and multiplicities - CATs - Educational Studies


Zeros and multiplicities << prev chpt: ... The zeros of a polynomial function p(x) ... Find the zeros and multiplicity of each polynomial.
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IMPORTANT! Find all the zeros of each function. ... - OpenStudy


State the multiplicity of any multiple zeros. y= ... Find all the zeros of each function. State the multiplicity of any multiple zeros. y= ...
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Suggested Questions And Answer :


Find zeros of the polynomial function & state the multiplicity of each. f(x) = 3(x + 8)2(x - 8)3

6*(x+8)^2 *(x-8)^3 zeroes at x=-8 (2 times) & at x=8 (3 times)
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find the zeros of the polynomial function and state the multiplicity of each f(x)= x^4(x-8)^2

x^4 ( x - 8 )^2 =  0 x^4 = 0 or ( x - 2)^2 = 0 x = 0 or x - 2 = 0 x = 0 or x = 8
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Find the zeros for the polynomial function and give the multiplicity for each zero

f(x) = 4(x^2)(x+6)^2 Find the zeroes:  x = 0, multiplicity 2, x = -6, multiplicity 2 Does the graph cross the x-axis? No.  Each part- 4, x^2, and (x+6)^2- can never be negative, so f(x) can never be negative. Does the graph touch the x-axis and turn around? In other words, does f(x) ever equal zero (without becoming negative)? Yes.  At x = 0 and x = -6.
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find the zeros of each function state the multiplicity of multiple zeros

start: y=x(x-8)^2 zeroes...y=0, 8, 8
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find a polynomial function of degree four with -3 as a zero multiplicity 1,

(x+3)(x-3)(x-3)(x+2) = 0 (x^2 - 9)(x^2 - x - 6) = 0 x^4 - x^3 - 6x^2 - 9x^2 + 9x + 54 = 0 x^4 - x^3 - 15x^2 + 9x + 54 = 0
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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.]
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find a polynomial function of degree 4 with -3

y=x(x+3)^3............
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Find a polynomial function to model the data.

polynomial function of degree4 with -4 as a zero of multiplicity 3 and 0 as a zero of multplicity 1
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Derivative Problem

To find the derivative, we will apply the chain rule. It states: if f(x) = g(h(x)) then f'(x) = g'(h(x))*h'(x) In other words, the derivative of f(x) is equal to the original inner function (h) plugged into the derivative of the outer function (g') multiplied by the derivative of the inner function (h'). You can consider f(x) = (2x^2-x+4)^3 to be a composite function. if f(x) = g(h(x)) then g(x) = x^3 [outer] and h(x) = 2x^2-x+4 [inner] Note: The "x" in g(x) is really h(x) as it is written out in f(x)'s full form (the original function). We'll need the derivatives of both g(x) and h(x). They can be easily differentiated using the power rule. g'(x) = 3x^2 h'(x) = 4x-1 Now, we apply the chain rule: f'(x) = 3(2x^2-x+4)^2 * (4x-1) Simplifying, we get f'(x) = 48x^5 - 60x^4 + 216x^3 - 147x^2 + 216x - 48 Having finished the problem, I think it might be faster to expand (2x^2-x+4)^3 and use the power rule on the expansion... Either way, this problem requires an excessive amount of polynomial multiplication. Tell your teacher to give more reasonable examples!
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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!
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