Find the relative minimum point(s) of

f(x)=x^4/4 - x^3 -5x^2 - 10

I know you have to take the derivative but I am not sure how to do that for this problem.

f'=x^3-3x^2-10x=0

x(x^2-3x-10)=0
x(x-5)(x+2)=0
x=0;5;-2

Now you have to test each to see if it is a minimum
f"=3x^2-6x-10
at x=0, it is negative, so it is a max
at x=5, 75-30-10 is positive, a relative min
at x=-2
f"=12+12-10, a rel minimum

oops, at x=-2, f" is negative, so a relative max.

To find the relative minimum point(s) of a function, you need to find the critical points and analyze their nature. To do this, you can follow these steps:

Step 1: Take the derivative of the function. In this case, you need to find the derivative of f(x)=x^4/4 - x^3 - 5x^2 - 10.

The derivative of x^n (where n is a constant) can be found using the power rule. The power rule states that the derivative of x^n is equal to n*x^(n-1). Applying this rule to each term of the function, we get:

f'(x) = (4/4)x^(4/4-1) - 3x^(3-1) - 5(2)x^(2-1) - 0,

Simplifying this expression, we have:
f'(x) = x^3 - 3x^2 - 10x.

Step 2: Set the derivative f'(x) equal to zero to find the critical points. In other words, solve the equation x^3 - 3x^2 - 10x = 0.

Step 3: Solve the equation x^3 - 3x^2 - 10x = 0 to find the critical points. You can either solve this equation analytically or by using numerical methods such as factoring, completing the square, or using a graphing calculator. Once you solve the equation, you will find the values of x that correspond to the critical points.

Step 4: After finding the critical points, determine the nature of each point to identify if it is a relative minimum. You can do this by using the second derivative test or by analyzing the behavior of the function on either side of the critical point.

Applying these steps, you should be able to find the relative minimum point(s) of the given function f(x)=x^4/4 - x^3 - 5x^2 - 10.