Transcript 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, f x  f a
 
 
xa
fail to approach a limit as x approaches a. Four instances
where this occurs:
1. A corner, where the one-sided derivatives differ.
Example:
f  x  x
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, f x  f a
 
 
xa
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:
f  x  x
23
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, f x  f a
 
 
xa
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:
f  x  x
3
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, f x  f a
 
 
xa
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 onesided derivatives to be nonexistent).
Example: The Unit Step Function
1, x  0
U  x  
1, x  0
There is a discontinuity at x = 0
Relating Differentiability and Continuity
Theorem: If f has a derivative at x = a, then
continuous at x = a.
f
is
Intermediate Value Theorem for Derivatives
If a and b are any two points in an interval on which
is differentiable, then f  takes on every value
between f  a and f  b .
 
 
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?
f
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?
f  x   x 1
g  x   x 2  0.0001  0.99
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?
f  x   x 1
g  x   x 2  0.0001  0.99
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?
f  x   x 1
g  x   x 2  0.0001  0.99
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…
Differentiability Implies Local Linearity
Is either of these functions differentiable at x = 0?
f  x   x 1
g  x   x 2  0.0001  0.99
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
Find all the points in the domain of
the function is not differentiable.
f  x   x  2  3 where
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.
 2,3
Guided Practice
For the given function, compare the right-hand and left-hand
derivatives to show that it is not differentiable at point P.
2, x  1 Left-hand derivative:
f  x  
f 1  h   f 1
22
2
x
,
x

1

lim
 lim
h 0
h 0
h
h
P 1, 2 
 lim 0  0
Graph the function:
P 1, 2 
Right-hand derivative:
h 0
f 1  h   f 1
lim
h 0
h
2 1  h   2
 lim
 lim 2  2
h 0
h 0
h

Guided Practice
For the given function, compare the right-hand and left-hand
derivatives to show that it is not differentiable at point P.
2, x  1
f  x  
2 x, x  1
P 1, 2 
Left-hand derivative: 0
Graph the function:
P 1, 2 
Right-hand derivative: 2
Since 0  2 , the function is
not differentiable at point P.
Guided Practice
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?
y  f  x  D : 2  x  3
(a) All points in [–2,3]
except x = –1, 0, 2
(b) x = –1
(c) x = 0, x = 2
Guided Practice
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.
45
Check the one-sided derivatives!!!
yx
y  0  h   y  0
lim
h 0

h
45
1
h
 lim 1 5
 lim
h 0 h
h 0
h
45
y  0  h   y  0
1
h
 lim 1 5
lim
 lim
h 0 h
h 0
h 0
h
h
The problem is a cusp!!!
(support with a graph???)
 

Guided Practice
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.
13
Check the two-sided derivative!!!
y  3 x
y  0  h   y  0
lim
h 0
h
3 h  3

 lim
13
h0
h
 h1 3 
 1 
 lim  
 lim   2 3 

h0
h 0
h
 h 


 
The problem is a vertical tangent!!!
(support with a graph???)