Check exactly one of the three check boxes above to select one of three functions. All three functions include the point (1, -3), and all three are discontinuous at x = 1 with different types of discontinuities.
Recall that for a function to be differentiable at the point x = a the following limit must exist:[br][math]f'\left(a\right)=lim_{h\longrightarrow0}\frac{f\left(a+h\right)-f\left(a\right)}{h}[/math]. For the derivative (this limit) to even have a hope of existing then the difference quotient (the fraction without the limit) must exist, and this can only happen if f(a) is defined. [br]Therefore, a function must exist at [i]x[/i] = [i]a[/i] for [i]f[/i] '([i]a[/i]) to exist.
Recall that a function is continuous if and only if [math]lim_{x\longrightarrow z}f\left(x\right)=f\left(a\right)[/math]. This means the function must be defined, the limit must exist, and they must be the same for the function to be continuous.[br][br]We know that the function must exist at x = a for both continuity and differentiablity at the point. Let's investigate what happens when the function is defined at x = a, but the function is not continuous there.[br][br]Recall that the difference quotient, [math]\frac{f\left(a+h\right)-f\left(a\right)}{h}[/math], is the slope of the secant line from the point of interest (a, f(a)) to a nearby point, B.[br][br]In the app: [br] Notice that the function graphed in blue is a piecewise function which has a discontinuity at x = 1. It is defined there so that (a, f(a)) = (1, -3). Move B around on the graph of f and see what happens to the difference quotient (slope of the red secant line).
Does the derivative exist at x = 1 for this function?
No, the derivative does not exist at this point.[br]Note that although the limit from the left exists for function 1 (but not for functions 2 and 3),[br]the limit from the right does not exist for any of the three examples since it approaches infinity[br]since the tangent line is approaching a vertical line.[br]Therefore, the derivative is undefined at x = 1.
Does a function have to be continuous at a point in order to be differentiable there?
Yes. For any function where the function exists at x = a, but x = a is a[br] discontinuity, the same problem will exist. The left or right limits [br]of the difference quotient will be infinity or negative infinity (i.e. [br]nonexistent) for one or both sides, making the full limit defining the [br]derivative undefined. A function must be continuous at a point in order[br] to be continuous there, i.e. differentiability implies continuity. [br]Therefore, the set of functions which are differentiable at x = a is a [br]subset of the set of functions which are continuous at x = a which is a[br] proper subset of the set of functions which are defined at x = a. [br][br]What about the other direction? If a function is continuous at x = a, [br]then does it have to be differentiable at x = a? I.e. does continuity [br]imply differentiablitiy? To investigate this see Dr. Jackson's GeoGebra[br] activity "Does Continuity Imply Differentiabiltiy?".