Consider the following function f(x) %3 (х — 3)3 (a) find the intervals where the function fis increasing and where is decreasing. (enter your answers using interval notation. an answer cannot be expressed as an interval, enter empty or .) increasing decreasing (b) find the relative extrema of f. (if an answer does not exist, enter dne.) relative maximum (х, у) - (х, у) %3d relative minimum (c) find the intervals where the graph of f is concave upward and where it is concave downward. (enter your answers using interval notation. if an answen cannot be expressed as an interval, enter empty or .) concave upward concave downward (d) find the inflection points, if any, of f. (if an answer does not exist, enter dne.) (х, у) —

can you rephrase the question ? it is unclear : -)

step-by-step explanation:

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! : )

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step-by-step explanation:

for volume, add base x height x width

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