I think you might mean f(x)+2? Or do you mean f(x+2)? Either way it depends on what f(x) is.
They are x = 0, -5 and +8.
The function is F(x)= x^3+3x^2-6x+20
h(x)=3x-4
f(x) = 2x + 1 g(x) = x^2 - 7 So f*g(x) = f(g(x)) = f(x^2 - 7) = 2*(x^2 - 7) + 1 = 2*x^2 - 14 + 1 = 2*x^2 - 13
what is the sum of
-2, 1.74 and 0.46
x^2 + 11x + 6 has no rational zeros.
They are x = 0, -5 and +8.
f(2) = 8 - 26 + 18 = 0 (x-2) is a factor of f(x), x = 2 f(x) = (x-2)(x2 + 2x -9) x2 + 2x - 9 = 0 (x+1)2 - 10 = 0 x + 1 = +-sqrt10 x = -1 + sqrt10, -1 - sqrt10, 2
No. Consider the polynomial: f(x) = x3 + 4x2 + 4x + 16 then f'(x) = 3x2 + 8x + 4 = (3x + 2)(x + 2) => x = -2/3, -2 are the zeros of f'(x) Using the second derivative: f''(x) = 6x + 8 it can be seen that: f''(-2) = -4 -> x = -2 is a maximum f''(-2/3) = +4 -> x = -2/3 is a minimum But plugging back into the original polynomial: f(-2) = 16 f(-2/3) = 14 22/27 Between the zeros of the first derivative, the slope of the polynomial is negative so that the polynomial is always decreasing in value, but as the polynomial is greater than zero at the zeros of the first derivative, it cannot become zero between them. That is it has no zeros between the zeros of its first derivative f(x) = x3 + 4x2 + 4x + 16 = (x + 4)(x2 + 4) has only 1 zero at x = -4.
The function is F(x)= x^3+3x^2-6x+20
The domain of the function f (x) = square root of (x - 2) plus 4 is Domain [2, ∞)
It is f(x) = -x2 or (-x)2, whichever you intended.
The zeros of f(x), a function of the variable x, are those values of x for which f(x) = 0. These are points at which the graph of f(x) crosses (or touches) the x-axis. Many functions will do so several times over the relevant domain and the values (of x) are the distinct zeros.
The answer is indeterminate because the degree of f(x) is not specified.
Each factor will contribute a zero. f(x) = Ax^2 + Bx + C will take the form of something like f(x) = (x-r1)(x-r2) when you factor it. Now the zeros are the values of x for which f(x) = 0. You know 0 times anything is zero, so consider one factor at a time. (x-r1) will equal zero when x = r1, therefore r1 is a "zero" (or "root") of f(x). Incidentally, f(x) = 0 when x = r1 because 0*(x-r2) = 0 for any value of r2. To find the second "zero" find the value of x that makes (x-r2) equal zero. If f(x) = (x+2)(x+3) then the zeros are -2 and -3, because f(-2) = 0 and f(-3) = 0. So if you can factor a function, you can easily find its zeros. The challenge is actually factoring the function.
F(x)=[x^2]+1