Mark H. answered • 12/18/19

Experienced Tutor Specializing in Algebra, Geometry, and Calculus

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 Nx^{n} yields a derivative "n*Nx^{n-1}.

Given f(x) = 7/12*x^{3} -9/4*x^{2} -6x, the critical point denoted as x_{o} satisfied the following relation:

f'(x_{o}) = 3*(7/12)*x_{o}^{2} -2*(9/4)*x_{o} - 6 =0 --> (21/12)*x_{o}^{2 }- (54/12)*x_{o }- (72/12) = 0--> (21*x_{o}^{2 }- 54*x_{o }- 72)/12 =0.

To solve x_{o}, we have x_{o }= (54 +/- √( 54^{2} -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)**