Differentiable
How to determine whether a function is differentiable: definition, 2 examples, and their solutions.
Definition
y = f(x) is differentiable at a point when
the left-hand derivative and the right-hand derivative are equal.
This is true because
the limit value is defined
when the left-hand limit and the right-hand limit are equal.
One-Sided Limits
In plain words,
y = f(x) is differentiable
when its graph changes smoothly:
connected continuously
and without any sharp points.
Example
Before determining whether
y = f(x) is differentiable,
first show that y = f(x) is continuous.
Find the left-hand limit of f(x)
at x = 1.
f(x) = x2 + 2 at x = 1-.
One-sided Limits
So limx → 1- f(x) = limx → 1- (x2 + 2).
This is equal to 3.
Find the right-hand limit of f(x)
at x = 1.
f(x) = -x2 + 4x at x = 1+.
So limx → 1+ f(x) = limx → 1+ (-x2 + 4x).
This is equal to 3.
Find f(1).
f(x) = -x2 + 4x at x = 1.
So f(1) = -12 + 4⋅1 = 3.
The left-hand limit is 3.
The right hand limit is 3.
And f(1) = 3.
The left-hand limit, the right-hand limit,
and the funtion value of f(x)
at x = 1
are all equal.
So f(x) is continuous at x = 1.
Next, determine whether
f(x) is differentiable at x = 1.
Find the left-hand derivative of f(x)
at x = 1.
limx → 1- f'(x) = limh → 0 [f(1- + h) - f(1-)]/h.
Derivative: Definition
If you learned the derivative formula,
you can use the formula in this step.
Derivative of a Polynomial
f(x) = x2 + 2 at x = 1-.
So f(1- + h) = (1- + h)2 + 2.
And f(1-) = (1-)2 + 2.
(1- + h)2
= (1 + h)2
= 1 + 2⋅1⋅h + h2
= 1 + 2h + h2
Square of a Sum
-[(1-)2 + 2] = -[1 + 2] = -1 - 2
Cancel 1 and -1.
Cancel +2 and -2.
Then limh → 0 [h2 + 2h]/h.
Divide both of the numerator and the denominator
by h.
Then limh → 0 [h + 2]/1.
Then limh → 0 [h + 2]/1 = [0 + 2]/1 = 2.
So the left-hand derivative of f(x) at x = 1
is 2.
Next, find the right-hand derivative of f(x)
at x = 1.
limx → 1+ f'(x) =limh → 0 [f(1+ + h) - f(1+)]/h.
f(x) = -x2 + 4x at x = 1+.
So f(1+ + h) = -(1+ + h)2 + 4(1+ + h).
And f(1+) = -(1+)2 + 4⋅(1+).
-(1+ + h)2
= -(1 + h)2
= -(1 + 2⋅1⋅h + h2)
= -(1 + 2h + h2)
+4(1+ + h) = +4 + 4h
-[-(1+)2 + 4(1+)] = -[-1 + 4]
-(1 + 2h + h2) = -1 - 2h - h2
-[-1 + 4] = +1 - 4
Cancel -1 and +1.
Cancel +4 and -4.
-2h + 4h = +2h
Then limh → 0 [-h2 + 2h]/h.
Divide both of the numerator and the denominator
by h.
Then limh → 0 [-h + 2]/1.
Then limh → 0 [-h + 2]/1 = [0 + 2]/1 = 2.
So the right-hand derivative of f(x) at x = 1
is 2.
The left-hand derivative is 2.
The right-hand derivative is 2.
The left-hand derivative
and the right-hand derivative
at x = 1
are equal.
So f(x) is differentiable.
So f(x) is differentiable
at x = 1.
This curve is the graph of y = f(x).
The graph is continuously connected
at x = 1.
So f(x) is continuous at x = 1.
The left-hand slope
and the right-hand slope
at x = 1
are both 2.
So the slope of y = f(x) exist.
So y = f(x) is differentiable at x = 1.
Example
Write f(x) = |x| as a piecewise function.
First show that y = f(x) is continuous.
Find the left-hand limit of f(x) at x = 0.
f(x) = -x at x = 0-.
So limx → 0- f(x) = limx → 0- -x.
This is equal to 0.
Find the right-hand limit of f(x) at x = 0.
f(x) = x at x = 0+.
So limx → 0+ f(x) = limx → 0+ x.
This is equal to 0.
Find f(0).
f(x) = x at x = 0.
So f(0) = 0.
The left-hand limit is 0.
The right hand limit is 0.
And f(0) = 0.
The left-hand limit, the right-hand limit,
and the funtion value of f(x)
at x = 0
are all equal.
So f(x) is continuous at x = 0.
Next, determine whether
f(x) is differentiable at x = 0.
Find the left-hand derivative of f(x)
at x = 0.
limx → 0- f'(x) = limh → 0 [f(0- + h) - f(0-)]/h.
f(x) = -x at x = 0-.
So f(0- + h) = -(0- + h).
And f(0-) = (-0-).
Solve the limit.
Then the limit value is -1.
So the left-hand derivative of f(x) at x = 0
is -1.
Next, find the right-hand derivative of f(x)
at x = 0.
limx → 0+ f'(x) = limh → 0 [f(0+ + h) - f(0+)]/h.
f(x) = x at x = 0+.
So f(0+ + h) = 0+ + h.
And f(0+) = 0+.
Solve the limit.
Then the limit value is 1.
So the right-hand derivative of f(x) at x = 0
is 1.
The left-hand derivative is -1.
The right-hand derivative is 1.
The left-hand derivative
and the right-hand derivative
at x = 1
are not equal.
So f(x) is not differentiable.
So f(x) is not differentiable
at x = 0.
This is the graph of y = f(x): y = |x|.
Absolute Value Function: Graph
The graph is connected continuously at x = 0.
So f(x) is continuous at x = 0.
The the left-hand slope at x = 0 is -1.
And the right-hand slope at x = 0 is 1.
The left-hand slope and the right-hand slope are not equal.
So there's a sharp point at x = 0.
So the slope of y = f(x) doesn't exist.
So y = f(x) is not differentiable at x = 0.