Identify the critical points of the function. & Determine the intervals on which the function increases and decreases.

Question

answer each of the following questions.

i. Identify the critical points of the function.

ii. Determine the intervals on which the function increases and decreases.

Answer 1

i. The critical points of a function are defined such that f'(x_o) = 0. For a polynomial function with constant N, the nth degree term, written as Nxⁿ yields a derivative "n*Nx^n-1.

Given f(x) = 7/12*x³ -9/4*x² -6x, the critical point denoted as x_o satisfied the following relation:

f'(x_o) = 3*(7/12)*x_o² -2*(9/4)*x_o - 6 =0 --> (21/12)*x_o² - (54/12)*x_o - (72/12) = 0--> (21*x_o² - 54*x_o - 72)/12 =0.

To solve x_o, we have x_o = (54 +/- √( 54² -4*21*-72))/(2*21)--> x_o = 1.28 +/- 2.254 --> x_o = 3.534 or -0.974.

ii To solve the points at which the function is increasing or decreasing, you'll need to find the points of inflection, which is yielded by taking the 2nd derivative of the function at the critical points. The inflection point is where the function changes concavity. A negative result implies the function is concave down whereas a positive result implies the function is concave up. The 2nd derivative determines how the slope of the function is changes with respect to x.

Therefore, f''(x) = f'(f'(x_o)) = (21*2*x_o -54)/12--> Using x_o = 3.534 yields f''(3.534) =(42*3.534 - 54)/12 = 7.86 >0 --> concave up or positive slope increase.

Similarly, f''(-.974) = (42*-.974 - 54)/12 = -7.9 < 0--> concave down or negative slope decrease

Therefore, the function decreases along the interval (-00, -.974]U[-.974,3.534) and the function increases along the interval [3.534, 00)