Find all the zeros of the function.?
f(x) = x^4 + 41x^2 + 400 Find all the zeros of the function. (Enter your answers as a comma-separated list.) x= Write the polynomial as a product of linear factors. f(x)= f(x) = x4 − 12x3 + 37x2 − 12x + 36 Write the polynomial as a product of linear factors. f(x)= Use your factorization to determine the x-intercepts of the graph of the function. Use a graphing utility to verify that the real zeros are the only x-intercepts. (Enter your answers as a comma-separated list. If no x-intercepts exist, enter DNE.) x= Use synthetic division to show that x is a solution of the third-degree polynomial equation, and use the result to factor the polynomial completely. List all the real solutions of the equation. x^3 − 57^x − 56 = 0, x = −1 x = (smallest value) x = x = (largest value) Find a polynomial function f(x) with real coefficients that has the given zeros. −1, −1, 3 + 5i f(x)= Thank you!!
Find the real zeros of the function: f(x)=(3x-7)^2-10(3x+7)+25?
f(x) = (3x + 7)² - 10(3x + 7) + 25 = 0 you will be able to correctly be tempted to enhance this expression, yet that may no longer needed. further you additionally could make an errors and be much extra puzzled. permit z = 3x + 7, so the equation will become z² - 10z + 25 = 0. This quadratic polynomial components as (z – 5)² = 0 for this reason z = 5 is a double root, i.e., the only answer. Now return to x: 3x + 7 = 5, or x = (5 – 7)/3 = -2/3. you're able to examine this effect interior the unique equation, I did. yet in a distinctive thank you to circulate is to realize the unique expression is a ideal sq. development. (3x + 7)² - 10(3x + 7) + 25 = [(3x + 7) – 5]² = 0 will become [3x + 2]² = 0, which has a double root so 3x + 2 = 0, or x = -2/3 ProfRay
Find the zeros of a cosine function.?
1 - 3cos(2x) = 0 ==> 3cos(2x) = 1 ==> cos(2x) = 1/3 ==> 2x = arccos(1/3) ==> x = (arccos(1/3))/2 so solutions are (arccos(1/3))/2 + kpi, for integer k
How do I find the zeros of the polynomial function [math]f(x)=\dfrac{1}{2}x^{3}-3x[/math]?
it will be :0=.5(x^3)-3x0=x^3–6x0=x(x^2–6)x=0 or x=6^.5x=-6^.5
How to find complex zeros and write the function in factored form?
one thank you to sparkling up a cubic is to seek for integer suggestions. because of the fact the made of the roots of a cubic polynomial equals minus the consistent term, look into the climate of 26: a million, 2, 13, 26. x=a million does not artwork, yet x=2 does. So 2 is one root. If 2 is a root, then our polynomial may be written interior the type: (x-2)*( a quadratic). to locate the quadratic, do the long branch: (x^3-8x^2+25x-26)/ (x-2) = x^2 -6x +13. Now use the quadratic equation to sparkling up this: you get x = [ 6 +/- sqrt(-sixteen) ]/2 = 3 +/- 2i. So the three roots of the cubic are 2, 3+2i, 3-2i. subsequently, we are in a position to write the polynomial in factored type as: (x-2)*[x-(3+2i)]*[x-(3-2i)].
Find the zeros of the rational function (if any).?
For the best answers, search on this site https://shorturl.im/avU6H The zeros of Rational Function is where R(x) = 0, here x which makes the numerator zero(not the denominator) The holes of Rational Function is where x makes the DENOMINATOR zero (just denominator not numerator). So if you find the domain of a function, the zeros must exist in domain and the holes don't exist in the domain. R(x) = 3x/(x^2 - 2x) x^2 - 2x ≠ 0 x(x - 2) ≠ 0 x ≠ 0 , 2 Domain : All real numbers - {0 , 2} Now we want to find where x makes numerator zero : 3x = 0 x = 0 Since x = 0 doesn't exist in domain, we have no answer a zero for this function. And for holes: x^2 - 2x = 0 x(x - 2) = 0 x = 0 , x = 2 Since x = 2 exists in domain, R(x) has a hole at x = 2. ** R(x) = 3x/(x^2 - 2x) R(x) = 3x / x(x - 2) You can cancel x from denominator and numerator : R(x) = 3 / (x - 2) Zero : no answer for x Hole : where the denominator is zero : x - 2 = 0, x = 2
Find the zeros of the function by factoring? HELP!! PLEASE!!?
You factor the polynomial into the product of two (or more) simpler polynomials (e.g A * B). Then any value that makes EITHER A or B zero, makes the WHOLE polynomial zero. (Because A * 0 = 0 * B = 0). So, for example, the first one factors into: (x + 8) (3x - 2) So x = -8 makes the first term zero and x = 2/3 makes the second one zero, so these are the zeroes of the original function. (I.e. they will make the original function zero also.)
Find all the zeros of each polynomial function. Use the zeros to write f in a factored form using...?
I assume that when you speak of 'zeros' you mean the roots of the function f(x) ie the values of x for which f(x) = 0. The easiest way of locating these values is simply to plot f(x) graphically and see where the curve crosses the x-axis. If this is not available or allowed, there are however analytical substitutes. For polynomials of order n, that is, the highest power of the independent variable x whose coefficient is not zero is x^n, the maximum number of roots is n. For the function given, which is a quartic or n = 4, there will be a maximum of 4 roots. When n is even, the minimum number of (real) roots is zero; when it is odd, you are guaranteed one real root. If the coefficients of the polynomial are integers, the systematic way to look for roots is to use the rational root theorem. This states that, if p is an integer factor of the constant coefficient An, and q an integer factor of the first coefficient Ao, then the real roots will be included in the set of values ±p/q. For the polynomial given, An = -40 and has factors p = 1, 2, 4, 5, 8, 10, 20 and 40, while Ao = 3 so that q = 1 or 3. To test each possible root ±p/q, it is substituted into f(x); if the result is zero, that value is a genuine root. The larger values are quickly eliminated, as the higher order terms dominate. Carrying out this procedure with the quartic provided identifies the four roots as +1, -2, -4 and -5/3, that is, there is one positive root and three negative roots. This enables the factored form of f(x) to be written immediately as f(x) = (x - 1).(x + 2).(x + 4).(3.x + 5) To confirm that this is correct, we expand it again to see if it produces the original polynomial f(x) = (x² + x - 2).(3.x² + 17.x + 20) = 3.x^4 + (17 + 3).x³ + (20 +17 - 6).x² + (20 - 34).x - 40 = 3.x^4 + 20.x³ + 31.x² - 14.x - 40 If some of the roots are imaginary, the factors will contain a quadratic or higher order term containing these roots, and the coefficients of this will have to be established by division by the linear terms containing real roots.
What is the range of the function f(x) = (3sinx - 4cosx +3 )/ (3sinx - 4cosx +2)?
Let us first simplify the given function.[math]f(x) = \frac{3sinx - 4cosx +3}{3sinx - 4cosx +2}[/math][math]=\frac{(3sinx - 4cosx +2)+1}{3sinx - 4cosx +2}[/math][math]=1+\frac{1}{3sinx - 4cosx +2}[/math]Range of the function [math]a sinx + b cosx + c[/math] is [math][c-\sqrt{a^{2}+b^{2}}, c+\sqrt{a^{2}+b^{2}}][/math].Range of [math]3sinx - 4cosx +2[/math] = [math][-3, 7][/math]As the range has both negative and positive portions and the actual function involves it’s reciprocal, let us split this into the union of two disjoint sets to avoid confusion.I am considering the range to be [math][-3, 0) \cup (0, 7][/math]. Here, [math]0[/math] is excluded intentionally because it is not desired in the denominator.Case(i): Considering [math][-3, 0)[/math]Range of [math]\frac{1}{3sinx - 4cosx +2}[/math] = [math](-\infty, -\frac{1}{3}][/math]Range of [math]1+\frac{1}{3sinx - 4cosx +2}[/math] = [math](-\infty, \frac{2}{3}][/math]Case(ii): Considering [math](0, 7][/math]Range of [math]\frac{1}{3sinx - 4cosx +2}[/math] = [math][\frac{1}{7}, \infty)[/math]Range of [math]1+\frac{1}{3sinx - 4cosx +2}[/math] = [math][\frac{8}{7}, \infty)[/math][math] \therefore[/math] Range of [math]f(x)[/math] = [math](-\infty, \frac{2}{3}] \cup [\frac{8}{7}, \infty)[/math]Thank you :)
Find all zeroes of [math]f(x) = x^3 - 2x?[/math]
Since the zeros of a polynomial function are those values of x for which the value of the function f, f(x), equals zero, then we'll set f(x) equal to zero and then solve for the independent variable x as follows: f(x) = x^3 - 2x = 0. Now, factoring the right side of this equation, we get: x(x^2 - 2) = 0. Now, by the Zero Product Property of Real Numbers, we know that if a(b) = 0, then either a = 0 or b = 0 or both a and b equal zero; therefore, we'll set each factor equal to zero and then solve for x: x = 0 or x^2 - 2 = 0. The solution for the first equation is obviously zero. For second, equation, we have: x^2 = 2, and now, taking the square root of both sides, we have: x = +/- 2^(1/2). Since the values of x = 0, - (2^(1/2)) and 2^(1/2) cause the value of f(x) = 0, then these are the zeros for the original polynomial function f(x) = x^3 - 2x.