Find all relative extrema. Use the Second Derivative Test where applicable. (If an answer does not exist, enter DNE.) f(x) = x^3 − 3x^2 + 3

Answer 1

f(x) = x³ - 3x² + 3

There are no values of "x" where the function DNE

To find the critical points, take the derivative and set it equal to zero:

f '(x) = 3x² - 6x (using the power rule)

3x² - 6x =0

3x(x - 2) = 0

x = 0 and x = 2

There are no points where the derivative DNE.

So x = 0 and x = 2 MAY be extrema or they may be points of inflection.

Take the second derivative:

f ''(x) = 6x - 6

Set it equal to zero:

6x - 6 = 0

6x = 6

x = 1

So x = 1 will be a point where the concavity changes. This means that x = 0 and x = 2 are NOT points of inflection and therefore are extrema. Now, to find out if they are minimums or maximums, plug in x = 0 and x = 2 to the second derivative to determine the concavity:

f ''(0) = 6(0) - 6 = -6 meaning, since the value is negative, that the function is concave down at this point. This make x = 0 a local maximum.

f ''(2) = 6(2) - 6 = 6 meaning, since the value is positive, that the function is concave up at this point. This make x = 2 a local minimum.