Question

5. For f(x) = 1 - x? on (-2, 1], do the hypotheses and conclusion of Rolle's Theorem hold? 6. Explain why not all of the hypotheses for the Mean Value Theorem hold for f(x) = 1 - Ixl on (-1,2]. A 5 Jul 20, Doc 2.pdf 29 smu sum....pdf test 28 math su....pdf 24

Answer 1

(5) Hero = 1-8² on [-2,1 Since a polynomial function is everywhere Congnuous and di Herenga ne. Therefore tid, is continuous on [-2,, and differenballe on (-2,1). NOW, f(-2) = 1629?= 1-4 =-3 f(1) = 1 - 1? 1 - 1 o fe-2) = f(1) since f(-2) + f(1) :: Rolle's the onem is not applilable to fix)=1-x² on [24] . 6 tex) = 1- lxl.on [-127 We have, Hal 1 -X OL x 22 E11-2 I +2, -1 < x 20 Since Polynomial function is erenywhene continuous and diffenengane. Therefore fr) is continuous and diffenentable for all aco as well as for all aro encept possibly at x=0. so, considen the point azo. we love him ton) = lim (1+) = 1+0=1 Ger ano 220 xto Rto lim sin lim (I-1)=1-01 xro -) X 70 No nyot 820) = I - 1 :: limno- fra fea) - hiso-fonos from

Then, tring is continuous at a=o. Hence, fins is continuous on [1,2]. Now lim fing-fo) (LHD at 2=o) = nto 2-o Tinto =RLO & fo=1 lim G+2)-1 aro - lim No 2 r r lim noo Livro anto 1 and (RHD at 2=0) = limon fing-fo) nyot R? 1-X-1 2 lm •nto 6708 fo=1 nto R 2 lim 1-1-1 n0 r lim -7 Hto R 2 -3 lim -1 Ro [- Cinto 270 = -1 :(LHD at na 220) & (RHD at a= =0) This shows that tirs is not differentiate at 220 € (-1,2), This, the condition of derivability at each point of (-1,2) is not satisfied. Hence, Mean Value Theorem is not hoked for ton = 1-121 on [1, 2]