What does the Second Derivative Test tell you about the behavior of f at these critical numbers?

Question

From Part (a), we have two critical numbers: x = 0

 and x = 15/4.

At x = 0

, f  ---Select--- has a relative minimum value has a relative maximum value does not have a relative extremum value may or may not have a relative extremum value because  ---Select--- f '' does not change sign at x = 0 f '' changes sign from - to + at x = 0 f '' changes sign from + to - at x = 0 f '' (0) is positive f '' (0) is negative f '' (0) = 0 or DNE .

At x = 15/4

, f  ---Select--- has a relative minimum value has a relative maximum value does not have a relative extremum value may or may not have a relative extremum value because  ---Select--- f '' does not change sign at x = 15/4 f '' changes sign from - to + at x = 15/4 f '' changes sign from + to - at x = 15/4 f '' (15/4) is positive f '' (15/4) is negative f '' (15/4) = 0 or DNE .

Answer 1

The second derivative test shows us how rapidly the slope of function f(x) is changing at critical numbers x=0, x=15/4. The slope of a function f(x) is described by the first derivative f'x). If the second derivative f''(x) at value "x" is positive, that means the slope of f(x) is increasing in steepness. If the second derivative at value "x" is negative, that means the slope of f(x) is decreasing in steepness. If the second derivative of f(x) is 0 at value "x," then the slope is constant (a linear function). We can use the second derivative test to help us sketch a polynomial on paper or with the help of a graphing calculator.

If the slope of a function is changing, we know right away that f(x) is not a linear function. It is most likely a polynomial function that has curves when graphed. In physics, the second derivative of a position function is the acceleration function for a particle or its change in velocity. Please comment if I missed something in this explanation.