find the stationary points of the following function f(x)=x^3-3x^2-24x-7
The stationary points are where df/dx = 0
So solve
3x^2 -6x -24 = 0
which is equivalent to
x^2 -2x - 8 = 0
(x-4)(x+2) = 0
That tells you you the two x values where the function is zero.
To find the stationary points of a function, we need to find the values of x at which the derivative of the function is equal to zero.
Let's start by finding the derivative of the function f(x). The derivative of x^3 is 3x^2, the derivative of -3x^2 is -6x, and the derivative of -24x is -24. The derivative of a constant (in this case, -7) is zero.
So, the derivative of f(x) is f'(x) = 3x^2 - 6x - 24.
Now, we need to find the values of x for which f'(x) = 0.
Setting f'(x) equal to zero and solving the equation:
0 = 3x^2 - 6x - 24
We can simplify this quadratic equation by dividing through by 3:
0 = x^2 - 2x - 8
To solve this equation, we can factor it or use the quadratic formula.
Factoring the equation:
0 = (x - 4)(x + 2)
Setting each factor equal to zero:
x - 4 = 0 --> x = 4
x + 2 = 0 --> x = -2
Therefore, the stationary points of the function f(x) = x^3 - 3x^2 - 24x - 7 are (4, f(4)) and (-2, f(-2)).