Question

Graph of A continuous function fis defined on the closed interval - 4sxs6. The graph of consists of a line segment and a curve that is tangent to the x-axis at x-3, as shown in the figure above. On th interval Dexc6, the function fis twice differentiable, with f(x)>0. Is there a value of a -4sach, for which the Mean Value Theorem applied to the interval (a 6), guarantees a value ca cx6, at which f'(c) = ? Justify your answer.

Answer 1

Given Function's Frol) graph is N FI) (61) - -4 -3 -2 -1 1 2 3 4 5 6 F130 F(6)= 1 OWN Given F is continuous on [-416 and is truire differentiable on [0,0]. E Requirement for the mean value theorena (1) F is continuous on Laib] ..@ F is differentiable on (a,b) But given b= 6: But at X= o f has Cooder point, so F is not differentiable but on [016 ] f is differentian ! For aco, on interval [016] mean value theorem can be applied rar q=3 in pesticular celso we as per our requrement since we went f'()to IMG also we can apply Se CAP

But on [61.37 F is decreasing So F! (D)<o: . 1a=3 Now on [3,6 ] f is continuous and on (316) F is differentiable 50 By mean value theorem, These exists ĆE (3,6). iii. such that , F! (0)= (G)- F(3) . 6- 3 FICO = 1 - 0 Flcc) = ☆