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.