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

What are the critical values?

Where are the relative extrema, if they exist?

To find the critical values of a function, we need to find its derivative and set it equal to zero.

Taking the derivative of f(x), we have:
f'(x) = 3x^2 + 6x - 2

Setting f'(x) equal to zero, we get:
3x^2 + 6x - 2 = 0

To solve this quadratic equation, we can use either factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)

For our equation, a = 3, b = 6, and c = -2. Substituting these values into the quadratic formula, we have:
x = (-6 ± sqrt(6^2 - 4(3)(-2))) / (2(3))
x = (-6 ± sqrt(36 + 24)) / 6
x = (-6 ± sqrt(60)) / 6
x = (-6 ± sqrt(4 * 15)) / 6
x = (-6 ± 2sqrt(15)) / 6
x = -1 ± sqrt(15)/3

Therefore, the critical values of f(x) are -1 + sqrt(15)/3 and -1 - sqrt(15)/3.

To find the relative extrema, we need to determine whether these critical values correspond to a maximum or minimum. We can do this by analyzing the second derivative of the function.

Taking the derivative of f'(x), we have:
f''(x) = 6x + 6

Now we can plug in the critical values into f''(x) to determine their nature.

For x = -1 + sqrt(15)/3:
f''(-1 + sqrt(15)/3) = 6(-1 + sqrt(15)/3) + 6
= -6 + 2sqrt(15) + 6
= 2sqrt(15)

Since f''(-1 + sqrt(15)/3) is positive, the function has a relative minimum at x = -1 + sqrt(15)/3.

For x = -1 - sqrt(15)/3:
f''(-1 - sqrt(15)/3) = 6(-1 - sqrt(15)/3) + 6
= -6 - 2sqrt(15) + 6
= -2sqrt(15)

Since f''(-1 - sqrt(15)/3) is negative, the function has a relative maximum at x = -1 - sqrt(15)/3.

In summary,

The critical values are: -1 + sqrt(15)/3 and -1 - sqrt(15)/3.
The relative extrema exist at: (-1 + sqrt(15)/3, relative minimum) and (-1 - sqrt(15)/3, relative maximum).