let f(x)=x^3+3x^2-2x+4

What are the critical values?

Where are the relative extrema, if they exist?

critical values are values in the domain where f'(x) = 0

since f'(x) = 3x^2 + 6x - 1, the critical values would be (-3±2√3)/2

the extrema occur at the critical values.

oops. f' = 3x^2 + 6x - 2.

extrema at x=(-3±√15)/3

To find the critical values of a function, you need to determine where its derivative is either zero or undefined. Let's begin by finding the derivative of the function f(x):

f'(x) = 3x^2 + 6x - 2.

Now, we can set this derivative equal to zero and solve for x to find the values where the derivative is zero:

3x^2 + 6x - 2 = 0.

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a),

where a = 3, b = 6, and c = -2.

Using the quadratic formula, we find two solutions for x:

x = (-6 ± √(6^2 - 4(3)(-2))) / (2(3))
= (-6 ± √(36 + 24)) / 6
= (-6 ± √60) / 6
= (-6 ± 2√15) / 6
= -1 ± (√15) / 3.

Hence, the critical values of f(x) are x = -1 + (√15) / 3 and x = -1 - (√15) / 3.

To determine the relative extrema, we can evaluate the second derivative of f(x), denoted as f''(x):

f''(x) = 6x + 6.

Now, we need to check the concavity of the function at the critical values obtained above. If f''(x) is positive, the function is concave up, indicating a relative minimum. If f''(x) is negative, the function is concave down, indicating a relative maximum.

Substituting the critical values into f''(x):

f''(-1 + (√15) / 3) = 6(-1 + (√15) / 3) + 6
= -2 + 2√15 + 6
= 4 + 2√15.

f''(-1 - (√15) / 3) = 6(-1 - (√15) / 3) + 6
= -2 - 2√15 + 6
= 4 - 2√15.

Since both f''(-1 + (√15) / 3) and f''(-1 - (√15) / 3) are positive, it indicates that there are no relative extrema for the function f(x).

Therefore, the critical values of f(x) are x = -1 + (√15) / 3 and x = -1 - (√15) / 3, but there are no relative extrema.