Find the x coordinates of all relative extreme points of
f(x)=1/4x^4 + 2/3x^3- 3/2x^2 + 4
f'(x)=x^3 + 2x^2 - 3x = 0
(x^2 + 3)(x-1) = 0
x=-3 x=1 x=0 Is this correct?
I get when factoring
x(x^2+2x-3)
x(x+3)(x-1)=0
x=0;-3, 1 same end as you, but your work is wrong.
thank you
To find the x-coordinates of all relative extreme points of a function, you need to find the critical points where the derivative of the function is equal to zero or is undefined. In this case, you correctly found the derivative of the function f(x) as f'(x) = x^3 + 2x^2 - 3x.
To determine the critical points, you can set the derivative equal to zero and solve for x:
x^3 + 2x^2 - 3x = 0
Factoring out an x from the equation, we have:
x(x^2 + 2x - 3) = 0
Now we need to solve for x using the zero-product property: if a product of factors is equal to zero, then at least one of the factors must be zero.
Setting each factor equal to zero, we have two options:
1) x = 0
2) x^2 + 2x - 3 = 0
Using the quadratic formula or factoring, we find the roots of the quadratic equation x^2 + 2x - 3 = 0:
x = 1 and x = -3
So you correctly found the critical points x = 0, x = 1, and x = -3.
Therefore, the x-coordinates of all relative extreme points of the function f(x) = 1/4x^4 + 2/3x^3 - 3/2x^2 + 4 are x = 0, x = 1, and x = -3.