Question

Consider the equation below. (If an answer does not exist, enter DNE.) FX) - - 2 - 15x + 4 (a) Find the interval on which is increasing (Enter your answer using interval notation) Find the interval on which is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of local minimum value local maximum value (c) Find the inflection point (X,Y) - Find the interval on which is concave up. (Enter your answer using interval notation) Find the interval on which is concave down. (Enter your answer using interval notation)

Answer 1

solution - Given f(n) = x²-6n²-150+ 4 f(n) = 30² - 122-15 (as for increasing , fniro 3n²12 nasyon 3(n²_4n-5) 70 2 31n 50tn-5) so 3(-5) (0171220 [(-2,-1) 4 (5, as for decreasing, finoco (6) local minimum value at (x=5] local maximum value at x = -1 Explanations at nas, s'en changes its sign – tot. at naut, finl changes it's sige & to an

( f(n) - 60-12 for inflection point , 60-12-0 16 (-2)0 n-2-0 -1022. at nag, y=(2,36 (2) ² - 15 (2) + 4 ya 8-24-30 +4. - - си Hence, (ny) = (12, -42] fony - con came up, frisos concave down , fonico

(2) ca, (1, 3) U (4,67 (increasing (6) decreasing 10,10 (3,4)] co) concave upward (d) concave downward (2, u) u (4,6) " Inflection points (9,41 = |(2,3)