Which of the following function is a polynomial function?
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it will be choice (c)\n(x+1)(x-1)\nx^2+x-x-1\nx^2-1
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Let f(x) = 3x^4 - 25 + 5/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nBecause f(x) is a polynomial function (of degree 4) on x, it follows that the function f is continuous on R (the set of real numbers).  So, if we take the limit on the left (and right) of x, we will get the same value for both sides. \n\nThat is, the behavior on the left (Left-hand Limit) will be: as x approaches  c from the left, we get f(x) approaches f(c) = 3c^4 - 25 + 5/4. Similarly, the behavior on the right (Right-hand Limit) will be: as x approaches c from the right, we also get f(x) approaches f(c) = 3c^4 - 25 + 5/4. \n\nTherefore, both the left and right-hand limit yields to f(c), i.e. as x approaches c, we always obtain f(c).\n\nHence, both left and right-hand behavior of f are the same for all real valued x.
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INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
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I believe you use the remainder theorem for this, which states that the remainder when a polynomial f(x) is divided by (x-k) is\nf(k)\n\nThus the remainder would be f(-1) = 11
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f(x) = [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2) ]\n= 3(x - 2)/(x - 3) (x ≠2, -2)\n= [3(x - 3) + 3]/(x - 3)\n= 3 + 3/(x - 3)\n\n1. Lim x→2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→2 [3 + 3/(x - 3)]  \n= 0 \nand \nLim x→-2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→-2 [3 + 3/(x - 3)] \n= 13/5\nSo it has 'holes' at (2, 0) and (-2, 13/5)\n\n2. It is undefined when x = 3, 2 and -2\nSo domain is all x except x = 3, 2, -2\nRange all y except y = 0 and 13/5\n\n3. As x increases and decreases without bound \n3/(x-3) approaches 0\nso 3 + 3/(x-3) approaches 3\n\nie lim x → ±∞ [3 + 3/(x-3)] = 3\n\n4. When x < 3\nx - 3 < 0 \nSo 3 + 3/(x-3) < 3\n\nWhen x > 3\nx - 3 > 0 \nSo 3 + 3/(x-3) > 3\n\nSo for x > 3 the curve is above y = 3 and for x< 3 the curve is below y = 3\n\n5. When y = f(x) = 0\n3(x - 2)/(x - 3) = 0\nie x = 2\nSo f(x) cuts x-axis at x = 2\n\n6. When x = 0\nf(x) = 3*(-2)/(-3) = 2\nSo y-intercept is y = 2\n\n7. It is rotationally symmetrical about the point (3, 3) (order 2) except for the 'holes'.\n\nThat should get you a good start I think
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Which function, Subsitution, Linear COmbination, Graph?
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Domain lists all possible x values. A normal polynomial function has a domain of all real numbers, because EVERY x value works.\n\nYour task is to find x values that DON'T work...and since you cannot divide by zero, then any x in the denominator that makes x^4+1 equal to zero will not be part of the domain.\n\nSo, what do you know about x^4+1?\n\nIt is an even power, so even if you put a negative number in for x, x^4+1 will be positive (since adding one will only make it more positive). If x=0, that is still positive, because of the plus one.\n\nx^4+1 is ALWAYS positive for any x...therefore it is NEVER equal to zero. Therefore any x value works, and your answer is all real numbers!!
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Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
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In mathematics, a polynomial is an expression in which a finite number of constants and variables are combined using only addition, subtraction, multiplication, and non-negative whole number exponents (raising to a power).
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The remainder is -43 because by the remainder theorem, f(c) is the remainder when a polynomial is divided by x-c. In this case, c is -2, so plug it into the equation and the remainder is what you get out of it.\n\n2(-2)(-2)(-2)-5(-2)(-2)+3(-2)-1\n= -16-20-6-1 = -43
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I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
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I am guessing your function to be one of the following:\n(1) f(x) = ½x - 3 or (2) f(x) = -½x -3\nIn function (1):\nf(0) = ½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = ½(2) - 3 = 1 - 3 = -2 ... or... (2, -2)\n\nIn function (2):\nf(0) = -½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = -½(2) - 3 = (-1) - 3 = -4 ... or ... (2, -4)
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So, for a) & b) the two things that you should notice are:\n1 - The exponents inside the second polynomial alternate, meaning that as the powers of the x variable decrease, the powers of the y variable increase, both by increments of one.\n2 - The answer will only contain the variables x and y raised to the same power, which is one more than the leading exponent on the x variable.  \n\nSince (x + y)(x^2 - xy + y^2) = x^3 + y^3\n and  (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) = x^5 + y^5,\nthen for any given polynomial the pattern is written out...\n(x + y)(y^2n - x^(2n-1)y + x^(2n-2)y^2 + ... + x^2y^(2n-2) - xy^(2n-1) + y^2n = x^(2n+1) + y^(2n+1)
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First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
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1. False, a polynomial has no = sign.\n2. False, the log of zero (or any negative number) does not exist.
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Polynomial is a general term for any expression, including monomial, binomial, trinomial, etc.  Monomials have just one term with no + or -, such as x or 3x or 5x^6.  Binomials are two monomials added or subtracted, such as 3x + 5 or 5x^2 - 6x.  The degree of a polynomial is the highest exponent of any of its variables, assuming there is only one variable.  So for x^3 + 8x^2 + 3x - 4, the degree is 3.
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Polynomials are expressions involving x raised to a whole number power (exponent). Some examples are: \n\n\n\nIn this lesson we consider division of polynomials such as: \n and . \nThere are two ways to calculate a division of polynomials. One is long division and a second method is called synthetic division. Synthetic division can be used when the polynomial divisor such as x-2 has the highest power of x as 1 and the coefficient of x is also 1. Otherwise, such as , a long division process must be used. Both methods will be demonstrated. \n\nWe will now look at some examples showing how these two procedures are done