To find the extreme values of the function, we need to find the critical points.
To do this, we take the derivative of the function and set it equal to zero:
y' = 3x^2 - 4x - 4 = 0.
Next, we solve for x:
3x^2 - 4x - 4 = 0.
This equation can be factored as:
(3x + 2)(x - 2) = 0.
Setting each factor equal to zero gives us:
3x + 2 = 0 or x - 2 = 0.
Solving these equations, we find that:
x = -2/3 or x = 2.
So, the critical points occur at x = -2/3 and x = 2.
To determine the type of extreme value at each critical point, we can use the second derivative test.
Taking the derivative of y' = 3x^2 - 4x - 4, we get:
y'' = 6x - 4.
Substituting the critical points into the second derivative, we have:
y''(-2/3) = 6(-2/3) - 4 = -4 - 4 = -8,
y''(2) = 6(2) - 4 = 12 - 4 = 8.
Since y''(-2/3) is negative while y''(2) is positive, we can conclude that there is a local minimum at x = -2/3 and a local maximum at x = 2.
To find the corresponding y-values, we substitute these x-values back into the original function:
y(-2/3) = (-2/3)^3 - 2(-2/3)^2 - 4(-2/3) + 5 = -8/27 + 4/9 + 8/3 + 5 = 67/27,
y(2) = 2^3 - 2(2)^2 - 4(2) + 5 = 8 - 8 - 8 + 5 = -3.
Therefore, the extreme values of the function occur at (x, y) = (-2/3, 67/27) and (x, y) = (2, -3).
Find the values of x where the extreme values of the function y=x3−2x2−4x+5 occur.
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