Differentiability, Local Linearity Section 3.2a

How f (a) Might Fail to Exist A function will not have a derivative at a point P (a, f(a)) where the slopes of the secant lines, fail to approach a limit as x approaches a. Four instances where this occurs: 1. A corner, where the one-sided derivatives differ. Example: There is a corner at x = 0

How f (a) Might Fail to Exist A function will not have a derivative at a point P (a, f(a)) where the slopes of the secant lines, fail to approach a limit as x approaches a. Four instances where this occurs: 2. A cusp, where the slopes of the secant lines approach infinity from one side and negative infinity from the other. Example: There is a cusp at x = 0

How f (a) Might Fail to Exist A function will not have a derivative at a point P (a, f(a)) where the slopes of the secant lines, fail to approach a limit as x approaches a. Four instances where this occurs: 3. A vertical tangent, where the slopes of the secant lines approach either pos. or neg. infinity from both sides. Example: There is a vertical tangent at x = 0

How f (a) Might Fail to Exist A function will not have a derivative at a point P (a, f(a)) where the slopes of the secant lines, fail to approach a limit as x approaches a. Four instances where this occurs: 4. A discontinuity (which will cause one or both of the one- sided derivatives to be nonexistent). Example: The Unit Step Function There is a discontinuity at x = 0

Relating Differentiability and Continuity Theorem: If has a derivative at x = a, then is continuous at x = a. Intermediate Value Theorem for Derivatives If a and b are any two points in an interval on which is differentiable, then takes on every value between and . Ex: Does any function have the Unit Step Function as its derivative? NO!!! Choose some a < 0 and some b > 0. Then U(a) = –1 and U(b) = 1, but U does not take on any value between –1 and 1  can we see this graphically?

Differentiability Implies Local Linearity Locally linear function – a function that is differentiable at a closely resembles its own tangent line very close to a. Differentiable curves will “straighten out” when we zoom in on them at a point of differentiability…

Differentiability Implies Local Linearity Is either of these functions differentiable at x = 0? 1. We already know that f is not differentiable at x = 0; its graph has a corner there. Graph f and zoom in at the point (0,1) several times. Does the corner show signs of straightening out? Continued zooming in at the given point (assuming a square viewing window) always yields a graph with the exact same shape  there is never any “straightening.”

Differentiability Implies Local Linearity Is either of these functions differentiable at x = 0? 2. Now do the same thing with g. Does the graph of g show signs of straightening out? Try starting with the window [–0.0625, 0.0625] by [0.959, 1.041], and then zooming in on the point (0,1). Such zooming begins to reveal a smooth turning point!!!

Differentiability Implies Local Linearity Is either of these functions differentiable at x = 0? 3. How many zooms does it take before the graph of g looks exactly like a horizontal line? After about 4 or 5 zooms from our previous window, the graph of g looks just like a horizontal line.  This function has a horizontal tangent at x = 0, meaning that its derivative is equal to zero at x = 0…

How does all of this relate to our topic of local linearity??? Differentiability Implies Local Linearity Is either of these functions differentiable at x = 0? 4. Now graph f and g together in a standard square viewing window. They appear to be identical until you start zooming in. The differentiable function eventually straightens out, while the nondifferentiable function remains impressively unchanged. Try the window [–0.03125, 0.03125] by [0.9795, 1.0205] How does all of this relate to our topic of local linearity???

Guided Practice There is a corner at (2,3), so this function is not Find all the points in the domain of where the function is not differentiable. Think about this problem graphically. We have the graph of the absolute value function, translated right 2 and up 3: There is a corner at (2,3), so this function is not differentiable at x = 2.

Guided Practice For the given function, compare the right-hand and left-hand derivatives to show that it is not differentiable at point P. Left-hand derivative: Graph the function: Right-hand derivative:

Guided Practice Left-hand derivative: 0 Right-hand derivative: 2 For the given function, compare the right-hand and left-hand derivatives to show that it is not differentiable at point P. Left-hand derivative: 0 Graph the function: Right-hand derivative: 2 Since , the function is not differentiable at point P.

Guided Practice (a) All points in [–2,3] except x = –1, 0, 2 The graph of a function over a closed interval D is given below. At what points does the function appear to be (a) differentiable? (b) continuous but not differentiable? (c) neither continuous nor differentiable? (a) All points in [–2,3] except x = –1, 0, 2 (b) x = –1 (c) x = 0, x = 2

Guided Practice The problem is a cusp!!! The given function fails to be differentiable at x = 0. Tell whether the problem is a corner, a cusp, a vertical tangent, or a discontinuity. Support your answer analytically. Check the one-sided derivatives!!! The problem is a cusp!!! (support with a graph???)

Guided Practice The problem is a vertical tangent!!! The given function fails to be differentiable at x = 0. Tell whether the problem is a corner, a cusp, a vertical tangent, or a discontinuity. Support your answer analytically. Check the two-sided derivative!!! The problem is a vertical tangent!!! (support with a graph???)