Let f(x)=5−12x+x3. Find (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. (a) f is increasing on the interval(s) (-inf,-2), (2,inf) (b) f is decreasing on the interval(s) (-2,2) (c) f is concave up on the open interval(s) (0,inf) (d) f is concave down on the open interval(s) (-inf,0) (e) the x coordinate(s) of the points of inflection are 0 Notes: In the first four boxes, your answer should either be a single interval, such as [0,1), a comma separated list of intervals, such as (-inf, 2), (3,4], or the word "none". In the last box, your answer should be a comma separated list of x values or the word "none".

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

table 3 represents a direct variation function. : )