So, (1) Any line with positive slope through (2,1) meets the condition for example f (x) = 3(x − 2) 1 has f '(x) = 3 > 0 for all x ≠ 2 and also for x = 2 graph {y1=3 (x2) 1907, 919, 245, 31} Replace 3 with any positive number for another exampleF x = x ^ { 2 } 6 x 8 f x = x 2 − 6 x 8 The equation is in standard form The equation is in standard form xf=x^ {2}6x8 x f = x 2 − 6 x 8 Divide both sides by x Divide both sides by x \frac {xf} {x}=\frac {\left (x4\right)\left (x2\right)} {x} HSBC244 has shown a nice graph that has derivative #f'(3)=0# Here are couple of graphs of functions that satisfy the requirements, but are not differentiable at #3# #f(x) = abs(x3)5# is shown below graph{y = abs(x3)5 14, 25, 616, 1185} #f(x) = (x3)^(2/3) 5# is shown on the next graph
Solution Graph The Piecewise Function Show All Work F X X If X Amp 04 0 And 0 If 0 Lt X Lt 5 And 2x 10 If X Amp 05 5 State Whether The Function Is Continuous If It Is
