the removable discontinuity is at .- (a) 0; (b) 0; (c) 1; (d) 1; (e) f(3) Does Not Exist; (f) a jump discontinuity at x = −3; a
removable discontinuity at x = 3 and an essential discontinuity at x = 5.
(a) If we look at the graph, we can see that f(x) = 0.(b) If we look at the graph, we can see that f(x) = 0.(c) If we look at the graph, we can see that f(x) = 1.(d) If we look at the graph, we can see that f(x) = 1.(e) f(3) Does Not Exist.(f) There are three discontinuities: (1) a jump discontinuity at x = −3; (2) a removable
discontinuity at x = 3; and (3) an essential discontinuity at x = 5.SOLUTIONS TO PRACTICE PROBLEM SET 3
1. 5
We find the derivative of a function, f(x), using the definition of the derivative, which is: f′(x) =. Here f(x) = 5x and x = 3. This means that f(3) = 5(3) = 15 and f(3 + h)
= 5(3 + h) = 15 + 5h. If we now plug these into the definition of the derivative, we get f′(3) = = . This simplifies to f′(3) = = 5.If you noticed that the function is simply the equation of a line, then you would have seen that
the derivative is simply the slope of the line, which is 5 everywhere.- 4
We find the derivative of a function, f(x), using the definition of the derivative, which is: f′(3) =. Here f(x) = 4x and x = −8. This means that f (−8) = 4(−8) = −32 and f
(−8 + h) = 4(−8 + h) = −32 + 4h If we now plug these into the definition of the derivative, we