Question

8. [-/1 Points] DETAILS SCALCET8 4.3.AE.002.NVA MY NOTES ASK YOUR TEAC EXAMPLE 2 Find the local minimum and maximum values of the function below. f(x) = 3x4 - 1683 - 192X2 + 5 Video Example SOLUTION f'(x) = 12x3 - 48x2 – 384x = 12x(x - 8)(x + 4) From the chart Interval 12x X-8 X + 4 f'(x) f x < -4 decreasing on (-0,-4) -4 < x < 0 + increasing on (-4,0) 0<x< 8 decreasing on (0,8) X >8 + increasing on (8,0) + + + + + + is a local minimum value by we see that f'(x) changes from negative to positive at -4, so f(-4) = the First Derivative Test. Similarly, f'(x) changes from negative to positive at 8, so f(8) = is a local minimum value. And, f(0) = is a local maximum value because f'(x) changes from positive to negative at 0. Need Help? Talk to a Tutor

Answer 1

4 3 We have, fo)= 331 -1671- 192:+5 to hax We have to find the critical points of f(x), Thus difti fox with respect to u, we e get a fon} = 1 2 37 - 68 - 192.x² + 5} t' (r) = 3.2 (17) - 16.0 (613) - 192.0 -672) 6) 3.x 423 – 16x3x²_192x22 to FGM) nux) du du d da = 1223_480² - 3892 = 122 (742 - 32 ) =124(x-8) fu+ 4) Now, for critical points of f(x), f), we have t')=0 eie. 12x (x = 4y - 32 ) = 0 (3242- 32) = 0 2.61-8) 62 +4 ) = 0 X-8=0, 4+9=0 거 or, By Middle term factor x=0 Method eil. = 8 x = - 4
The answer sheet has three pages.it is the first page
critical points at This fon has three and 28 x = -4, n=0 have to Now near all these check the sign of fox) critical points of f(n). The refore, the estervas 2,-4), (-4,0), (0, 8) and (8,4). monotone are 누 Interal 12u n- -8 +4 t'en decreasing on X <- 4 - de E-4) -4<H20 increasing on (x,-4) 64,5) on (0,8) - t 1 t decreasing OCUL 8 + - increasing on (ره) به 278 + + + When A function f is increasing when f'oyo and decreasing t'a co Thus we see from the chart that t') changes from negative to positive at-4 So, t+4)= 3. (-4) – 16.64) 3 192.64)² +5 4. 768 + 10 24 - 3072 +5 U 1797 - 3072 1275
second page
So f(-4)= 1275 local minimum Test. value by the first derivative Now, f (1) changes from negative to positive so f (8) 4 3.4 (8) 3 16. (8) 1928 82 +5 = 12288 – 3192 - 12.264+5 - 8187 8187 So, 78) - es local minimum value by first derivative Testi and FO) = 3 x 0 - 16x0 - 192X0 +5 - = 5 and f(o) = 5 és Value because local maximem f'60 changes positive to negative at x = 0
Third/last page