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
To find the zeros of the function ( f(x) = x\sqrt{2} + 4x + 8 ), we set the equation equal to zero: ( x\sqrt{2} + 4x + 8 = 0 ). This can be rewritten as ( x(\sqrt{2} + 4) + 8 = 0 ). Solving for ( x ), we get ( x(\sqrt{2} + 4) = -8 ), leading to ( x = \frac{-8}{\sqrt{2} + 4} ). Thus, the zeros of the function are ( x = \frac{-8}{\sqrt{2} + 4} ).
The function ( f(x) = x^2 - 6x + 8 ) is a polynomial function because it is a quadratic expression. To find the zeros, we can factor it as ( (x - 2)(x - 4) ), which gives us the zeros ( x = 2 ) and ( x = 4 ). Thus, the zeros of the function are 2 and 4.
-2, 1.74 and 0.46
x^2 + 11x + 6 has no rational zeros.
To find the zeros of the polynomial function ( f(x) = x^3 - 2x^2 - 8x ), we first factor the expression. We can factor out ( x ) from the polynomial, giving us ( f(x) = x(x^2 - 2x - 8) ). Next, we can factor the quadratic ( x^2 - 2x - 8 ) into ( (x - 4)(x + 2) ), leading to ( f(x) = x(x - 4)(x + 2) ). Therefore, the zeros of the function are ( x = 0 ), ( x = 4 ), and ( x = -2 ).
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
To find the quadratic polynomial whose zeros are 2 and -3, we can use the fact that a polynomial can be expressed in factored form as ( f(x) = a(x - r_1)(x - r_2) ), where ( r_1 ) and ( r_2 ) are the zeros. Here, substituting ( r_1 = 2 ) and ( r_2 = -3 ), we have ( f(x) = a(x - 2)(x + 3) ). Expanding this, we get ( f(x) = a(x^2 + x - 6) ). For simplicity, we can choose ( a = 1 ), giving us the polynomial ( f(x) = x^2 + x - 6 ).
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.
Yes, you can determine the zeros of the function ( f(x) = x^2 - 64 ) using a graph. The zeros correspond to the x-values where the graph intersects the x-axis. By plotting the function, you can see that it crosses the x-axis at ( x = 8 ) and ( x = -8 ), which are the zeros of the function.
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, ∞)