course mth 158
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23:26:15 4.2.20 (was 4.2.10). If f(x)= (x^2-5) / x^3 a polynomial?
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RESPONSE --> no
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23:26:37 This is not a polynomial function. It is the ratio of two polynomials, the ratio of x^2 - 5 to x^3.
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RESPONSE --> ok
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23:49:30 4.2.40 (was 4.2.30). If a polynomial has zeros at x = -4, 0, 2 the what is its minimum possible degree, and what form will the polynomial have?
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RESPONSE --> f(x)=a(x-(-4))(x-0)(x-2) for a=1 f(x)=(x+4)(x)(x-2) (x^2+4x)(x-2) x^3-2x^2+4x^2-8x x^3+2x^2-8x
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23:50:33 The factors of the polynomial must include (x - (-4) ) = x + 4, x - 0 = x, and x - 2. So the polynomial must be a multiply of (x+4)(x)(x-2). The general form of the polynomial is therefore f(x)=a(x+4)(x-0)(x-2).
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RESPONSE --> i solved it a little bit farther like the book shows i see
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00:07:28 4/2/52 (was 4.2.40). What are the zeros of the function f(x)=(x+sqrt(3))^2 (x-2)^4 and what is the multiplicity of each?
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RESPONSE --> the real zeros are -3with multiplicity of1 because sqrt 3would be 1/2 and then ^2 would be 1 and then the next zero is 2 with multiplicity of 4
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00:13:27 f(x) will be zero if x + sqrt(3) = 0 or if x - 2 = 0. The solutions to these equations are x = - sqrt(3) and x = 2. The zero at x = -sqrt(3) comes from (x + sqrt(3))^2 so has degree 2. The zero at x = 2 comes from (x-2)^4 so has degree 4.
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RESPONSE --> ok i see i went to far into the sqrt 3 . i could not find an example in the book about sqrt except for example 1 which showed g(x)=sqrt x which is showing it is raised to the 1/2 power so that is how i came up with my answer
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00:16:01 For each zero does the graph touch or cross the x axis?
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RESPONSE --> the graph crosses the x axis -sqrt3 and touches at 2
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00:16:11 In each case the zero is of even degree, so it just touches the x axis. Near x = -sqrt(3) the graph is nearly a constant multiple of (x+sqrt(3))^2. Near x = 2 the graph is nearly a constant multiple of (x - 2)^4.
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RESPONSE -->
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00:18:32 What power function does the graph of f resemble for large values of | x | ?
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RESPONSE --> function resembles y = x^6 for large values |x|
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00:18:38 If you multiply out all the terms you will be a polynomial with x^6 as the highest-power term, i.e., the 'leading term'. For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function x^6. **
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RESPONSE --> ok
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00:47:12 4.2.62 (was 4.2.50). f(x)= 5x(x-1)^3. Give the zeros, the multiplicity of each, the behavior of the function near each zero and the large-|x| behavior of the function.
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RESPONSE --> zeros are=1with multiplicity of 3 behavior of function=resembles that of the power function y=x^3 for large values of |x|
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00:50:59 The zeros occur when x = 0 and when x - 1 = 0, so the zeros are at x = 0 (multiplicity 1) and x = 1 (multiplicity 3). Each zero is of odd degree so the graph crosses the x axis at each. If you multiply out all the terms you will be a polynomial with 5 x^4 as the highest-power term, i.e., the 'leading term'. For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function 5 x^4.
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RESPONSE --> ok i see i do understand the zeros but the behaviour is still a little confusing
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00:53:32 What is the maximum number of turning points on the graph of f?
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RESPONSE --> the maximum # of turning points= n-1 3-1=2
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00:55:08 This is a polynomial of degree 4. A polynomial of degree n can have as many as n - 1 turning points. So this polynomial could possibly have as many as 4 - 1 = 3 turning points.
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RESPONSE --> i know exactly what i have done was to take the 5x(x-1)^3 exponent of 3 not the 5x^4 exponent and minus one i do understand
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00:59:16 Give the intervals on which the graph of f is above and below the x-axis
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RESPONSE --> (-infinity, 0)(0,1)(1,infinity)
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01:00:53 this polynomial has zeros at x = 0 and x = 1. So on each of the intervals (-infinity, 0), (0, 1) and (1, infinity) the polynomial will lie either wholly above or wholly below the x axis. If x is a very large negative or positive number this fourth-degree polynomial will be positive, so on (-infinity, 0) and (1, infinity) the graph lies above the x axis. On (0, 1) we can test any point in this interval. Testing x = .5 we find that 5x ( x-1)^3 = 5 * .5 ( .5 - 1)^3 = -.00625, which is negative. So the graph lies below the x axis on the interval (0, 1).
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RESPONSE --> (-infinity, 0)(0,1)(1,infinity) i believe i read the question wrong or the answer given is way into detail
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