Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x^
ID: 2874443 • Letter: C
Question
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x^3 - 3x^2 - 9x + 5 (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. local minimum value local maximum value (c) Find the inflection point. (x, y) = () Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)Explanation / Answer
f(x) = x3 - 3x2 - 9x + 5
f '(x) = 3x3-1 - 3(2)x2-1 - 9(1) + 0
==> f '(x) = 3x2 - 6x - 9
==> f '(x) = 3(x2 - 2x - 3) = 3(x + 1)(x - 3)
function is increasing ==> f '(x) > 0
==> 3(x + 1)(x - 3) > 0
==> (x + 1)(x - 3) > 0
==>x belongs to (- , -1) U (3 , )
function is increasing in the interval (- , -1) U (3 , )
function is decreasing ==> f '(x) < 0
==> 3(x + 1)(x - 3) < 0
==> (x + 1)(x - 3) < 0
==> x belongs to (-1 , 3)
==> function is decreasing in the interval (-1 , 3)
critical points ==> f '(x) = 0
==> 3(x + 1)(x - 3) = 0
==> x = -1 , 3
f '(x) = 3x2 - 6x - 9
f ''(x) = 3(2)x2-1 - 6(1) - 0
==> f ''(x) = 6x - 6
f ''(-1) = 6(-1) - 6 = -12 < 0
==> local maximum at x = -1
value = f(-1) = (-1)3 - 3(-1)2 - 9(-1) + 5 = 10
local maximum value = 10
f ''(3) = 6(3) - 6 = 12 > 0
==> local minimum at x = 3
value = f(3) = 33 - 3(3)2 - 9(3) + 5 = -22
local minimum value = -22
inflection point ==> f ''(x) = 0
==> 6x - 6 = 0
==> x = 1
f(1) = 13 - 3(1)2 - 9(1) + 5 = -6
Inflection point at (1 , -6)
concave up ==> f ''(x) > 0
==> 6x - 6 > 0
==> x > 1
==> x belongs to (1 , )
concave up in the interval (1 , )
concave down ==> f ''(x) < 0
==> 6x - 6 < 0
==> x < 1
Interval is (- , 1)
concave down in the interval (- , 1)