Find critical values of the function. If they exist, label as max or min / inflection point. Must show and use f' and f'' (use quadratic formula and round two decimal places). x^3 - 4x^2 - 5x + 7

Answer 1

Take the first derivative and set it equal to zero to find the critical values.

For f(x) = x³ - 4x² - 5x + 7, f '(x) = 3x² - 8x - 5

3x² - 8x - 5 = 0

Using the quadratic formula, x = [-(-8) ± √((-8)² - 4(3)(-5))]/(2•3) = (8 ± √76)/6 or x = 3.19 and x = -0.52

Since we are told to use f ''(x), we can do the second derivative test to see if these are minimum or maximum values.

f ''(x) = 6x - 8

If f ''(a) is positive, then the function is concave up at x = a meaning that (if "a" is a critical value) there would be a local minimum at x = a, and f ''(a) is negative, then the function is concave down at x = a meaning that there would be a local maximum at x = a.

f ''(3.19) = 6(3.19) - 8 = 11.14 so the graph is concave up meaning there is a local minimum at x = 3.19

f ''(-0.52) = 6(-0.52) - 8 = -11.14 so the graph is concave down meaning there is a local maximum at x = -0.52

Possible inflection points are found by setting the second derivation equal to zero so 6x - 8 = 0 when x = 4/3. Since we have already determined that to the right this point the graph is concave up and to the left it is concave down, then x = 4/3 is a point of inflection.