Many diabetics need to control their blood sugar levels with insulin. A graph is given that shows blood sugar levels over time after insulin is given to a non-diabetic person at time, t = 15 minutes. As you can see insulin causes blood sugar levels to drop. Where is the function not differentiable and why? Blood Sugar Levels Time 0 The function is differentiable everywhere except t = 30. O The function is differentiable everywhere. O The function is differentiable everywhere except t = 15, t = 30, and t = 40 where there are corners. O The function is differentiable everywhere except at t = 15 where there is a discontinuity. Water changes chemical structure as it is heated. It goes from ice, to liquid, to boiling. This is called a phase change. The graph given shows how the temperature of water changes as it is heated. You can identify phase changes on a graph by finding places where the change in temperature over time (derivative) is very small or zero. Where are the phase changes in the graph below? Temp 100 2:88 Temperature 10 15 20 25 30 35 40 45 50 Time O At times t = 35 to t = 45. O At times t = 15 to t = 35 and from t = 45 to t = 50. O At times t = 0 to t = 10 and from t = 25 to t = 35. O At times t = 25 to t = 30 and from t = 45 to t = 50.f (x) = lim f(xth )-f(x) h-0 h iff (x) = Ixl. Use the definition of the derivative to determine if the function is differentiable at x = 0. f (X) is not differentiable at x = 0 because the lim f (xth)-f(x) increases without bound at x = 0. h -c h o f (x) is not differentiable at x = 0 because the lim f(xth)-f(x) h -c h is not the same from the right and left. o f (x) is not differentiable at x = 0 because the lim f(xth)-f(x ) = 0 at x = 0. h-0 h o f (x) is differentiable at x = 0.\fWhere is the graph shown not differentiable and why? LY 5- -N X -10 -9 -8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 o f (x) is not differentiable at x = 0 because f (X ) is discontinuous at x = 0. o f (X) is not differentiable at x = 0 because f (X ) has a cusp at x = 0. o f (X) is differentiable at x = 0 because f (X ) is continuous at x = 0. o f (X) is not differentiable at x = 0 because I (X ) has a corner at x = 0.Where is the graph shown not differentiable and why? X -10 -9 -8 -7 -6 -5 -4 -3 2 -1 1 2 3 4 5 6 7 8 9 10 -1 + O f (X) is not differentiable at x = 1 because f (X ) is discontinuous at x = 1. o f (X) is not differentiable at x = 1 because I (X ) has a cusp at x = 1. o f (X) is not differentiable at x = 1 because I (X ) has a corner at x = 1. O f (X) is not differentiable at x = 1 because f (X ) has a vertical tangent at x = 1.Where is the graph shown not differentiable and why? Y -NowAUT -10 -9 -8 -7 -6 -5 -4 -3 -2 -1.1 1 3 4 5 6 7 8 9 10 o f (X) is not differentiable at x = 1 because I (X ) has a corner at x = 1. O f (X) is not differentiable at x = 1 because f (X ) is discontinuous at x = 1. o f (X) is differentiable everywhere shown. O f (X) is not differentiable at x = 1 because I (X ) has a cusp at x = 1.Which statement below is true? O A cusp is a point where the graph is not continuous. Q It is possible for a function to be differentiable but not continuous. Q If a graph is continuous then it must be differentiable. Q If a function is differentiable everywhere then it must be continuous. Why is the following function not differentiable at x = 1? O the left handed limit of f (X) does not exist at x =1? 0 f (X) is discontinuous at x =1. 0 f(X) has a cusp at x =1 0 f (X) has a corner at x =1 . The graph shown is continuous everywhere but not differentiable at many points. Why? -1U-8-8-?-8-5-4-3-2-10123458?8810 -1 -2 -3 -4 -5 O f (X) has a cusp and or corners/cusps. O f (X) has a vertical tangent line at x = O. O f (X) is differentiable everywhere 0 There are discontinuities in the graph