Question

B. 1-/1 Points DETAILS SESSCALC23.2.004 Verity that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers that still the conclusion of Role's Theorem. (Enter your answers as a comma-separated list) R) - 01/12, 7/12] 9. -17 Points DETAILS SESSCALC2 33.001.MI. Corder the equation below. (a) Pied the interval on which is increasing. (Enter your answer in interval notation) Find the interval on which is decreasing. (Enter your answer in interval notation) 0) Find the local minimum and mamum values of local minimum local maximum e Find the infection point sy- Find the interval on which concave up. (Enter your answer in interval notation) Find the interval on which is concave down. (Enter your answer in interval notation)

Answer 1

7T Given function f(x) = cos 3x, [ function Rolle's theoreni: het fi [a, b]-) R is which satisfied (i) fox) is continuous on [a,b] ii) differentiable on ra, b) fcal= f(b) then there exists Ceca, b) such that fee fccl=0 since On i) f(x) f(x)= cos3x, is trigonometric function is continuous [ 17/2 27] is differentiable on ii) f(x) - = COST 4 1) ( 7 ) = cos 3 1 / 2 va = COSTI 4 -f ( 727 ) = 10 Sp.: 1 = 60 *ZA - COSE = COS 27 = cos(-24-1) fo):117) hena +() salisyied three conditions of Poke'sthe way f(x) = -3 sinza {(c) = -3 sin3c f'(o)=0 =) -3 sin3c=0 let Sinac - sinna 3c=nT C = na 3

for na 77 € [11 YT 12 2 hence there experts c=1 [AA] such trat 1 s'co) = f (7) = 0 f(x) satisfied conditions of Rolle's theorem 9 f(x) = 2x² + 3x - 72x a) find the interval on which is increasing f'(x) = 2.3 x ² +3.2x - 72(1) f'cx) = 6x² + 6x-79 (x) = 6(x²+x-12) f(x) > 0 We know that fra) is incre creasing it 6(x²+x-12 ) > 0 2 x tx-19 o 2. x +4x-32-12 > 0 X(x+4)-3(x+4) 70 (x+4) (x-3) 70 . The interval in which f(x) is increasing is (-00,-4) U (3,0)

3 The to find the interval in which & is decreasing f(x) is decreasing if f(x) 10 (x+3)(x+4) <o (2-04))(x-3) < 0 interval is (-4,3) 6) find the ingle b) TO find the local manna and minimum &'(x)=0 (x+4) (x-3) = 0 x-3=0 x+4=0 x=-4 x=3 Now evaluate the sign of the first deuvative around each local extremum around f'(-5) = 6 (1-5) ² _5-18) = 6 (95-5-12) = 48-) ix reary $'(-3) = 6 (-3)?-3-12 ) = 6(9-3-12) 3-36–decrewing since the first derivative switches from increasing to decreasing there is a local maumum at x=-4. x = 3 - decker Around f'(a) = 6 (af 2-12 ) = 6 (448-18) = -36 f'(4) = 6 (4² - 12) = 48 - increaning

since the first derivative switches from decreasing to increasing there is a local minimum at x=3. local manunum = f(-4) = 2(-4)3 + 3 (-4) 278 (-4) 4 E-128 +48 728 8 local maximum = 208 local minimum = $(3)-& (3)3+3(3)2-7a(3) = 54 +97-316 local minimum --135 c) find the inflexion point plex)=0 6 (ant 1-0)=0 gato 2. up it f (x) > 0 f(x) is concave Ⓡx+ > 2x7-1 x-1 Solis concave up on the interval (- / 0) f(x) is concave down it f'(x) < 0 axt ico f(x) is concave down on the interval (- 00 ,- 1)