Finding relative extremes/extrama x-coordinates using the Second Deritavtive Test to find the x-coordinates of all relative extremes/extrema

Question

Hi, I got a question regarding using a Second Derivative Test to find the x-coordinates of all the relative extremes/extrema if they exist for f(x) = x⁴ - 4x³ + 4x² .

I took two derivatives of this and got 12x²-24x+8 or 4(3x² - 6x + 2) and set that equal to 0 and using the quadratic equation, and got (3+√3)/3 and (3-√3)/3 but I don't know if these are relevant to the problem to find the relative extremes, if they are, then I don't know/remember how to utilize them for this problem..

Answer 1

The 2nd derivative can be used for the "Test for Concavity" in which case you do set y'' equal to zero to determine "hypercritical" numbers.

But the 2nd derivative can also be used for the "Second Derivative Test" which allows you to determine if a critical number (found from the 1st derivative) is a relative max or min for the original function.

f(x) = x⁴ - 4x³ + 4x²

f'(x) = 4x³ - 12x² + 8x

f''(x) = 12x² - 24x + 8

Set the first derivative equal to zero to determine critical numbers (x-values where the slope of a tangent line is zero, for example).

4x(x² - 3x + 2) = 0

4x(x - 2)(x - 1) = 0

So, the critical numbers are x = 0, or x = 1, or x = 2.

At this point you could use the 1st derivative test to determine where the original function is increasing / decreasing which in turn would allow you to determine if those critical numbers are generating relative max or min for the original function.

But if you have gone to the trouble of finding the 2nd derivative (perhaps to use the Test for Concavity to determine points of inflection), you can also use the 2nd derivative to determine rel max/min for the list of critical numbers.

f''(x) = 12x² - 24x + 8

f''(0) = 8, which is positive, so original function is concave up (think positive, smiley face, up) at the point where x = 0. Since the tangent line has a slope of zero at that point, then (0, f(0)) is a relative min.

f''(1) = -4, which is negative, so original function is concave down (think negative, frown, down) at the point where x = 1. Since tangent line at that point has a slope of zero, the point (1, f(1)) is a relative max.

f''(2) = 8 (see f''(0) => relative min.

This graph will help visualize--visit the graph and take some notes:

desmos.com/calculator/29cc7gwfd9