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Precalculus
Introduction to calculus
Derivatives
The graph of
f
(
x
)
=
x
1
3
has a
vertical tangent at
x
=
0.
A General Note
Differentiability
A function
f
(
x
)
is differentiable at
x
=
a
if the derivative exists at
x
=
a
, which means that
f
′
(
a
)
exists.
There are four cases for which a function
f
(
x
)
is not differentiable at a point
x
=
a
.
When there is a discontinuity at
x
=
a
.
When there is a corner point at
x
=
a
.
When there is a cusp at
x
=
a
.
Any other time when there is a vertical tangent at
x
=
a
.
Determining where a function is continuous and differentiable from a graph
Using
[link] , determine where the function is
continuous
discontinuous
differentiable
not differentiable
At the points where the graph is discontinuous or not differentiable, state why.
The graph of
f
(
x
) is continuous on
(
−∞,
−2
)
∪
(
−2
, 1
)
∪
(
1,
∞
). The graph of
f
(
x
) has a removable discontinuity at
x
=
−2 and a jump discontinuity at
x
=
1. See
[link] .
Three intervals where the function is continuous
The graph of is differentiable on
(
−∞,
−2
)
∪
(
−2,
−1
)
∪
(
−1,
1
)
∪
(
1,
2
)
∪
(
2,
∞
)
. The graph of
f
(
x
) is not differentiable at
x
=
−2 because it is a point of discontinuity, at
x
=
−1 because of a sharp corner, at
x
=
1 because it is a point of discontinuity, and at
x
=
2 because of a sharp corner. See
[link] .
Five intervals where the function is differentiable Got questions? Get instant answers now! Got questions? Get instant answers now!
Try It
Determine where the function
y
=
f
(
x
)
shown in
[link] is continuous and differentiable from the graph.
The graph of
f
is continuous on
(
−
∞
,
1
)
∪
(
1
,
3
)
∪
(
3
,
∞
)
.
The graph of
f
is discontinuous at
x
=
1
and
x
=
3.
The graph of
f
is differentiable on
(
−
∞
,
1
)
∪
(
1
,
3
)
∪
(
3
,
∞
)
.
The graph of
f
is not differentiable at
x
=
1
and
x
=
3.
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Finding an equation of a line tangent to the graph of a function
The equation of a tangent line to a curve of the function
f
(
x
)
at
x
=
a
is derived from the point-slope form of a line,
y
=
m
(
x
−
x
1
)
+
y
1
.
The slope of the line is the slope of the curve at
x
=
a
and is therefore equal to
f
′
(
a
)
, the derivative of
f
(
x
)
at
x
=
a
.
The coordinate pair of the point on the line at
x
=
a
is
(
a
,
f
(
a
)
)
.
If we substitute into the point-slope form, we have
The equation of the tangent line is
y
=
f
'
(
a
)
(
x
−
a
)
+
f
(
a
)
A General Note
The equation of a line tangent to a curve of the function
f
The equation of a line tangent to the curve of a function
f
at a point
x
=
a
is
y
=
f
'
(
a
)
(
x
−
a
)
+
f
(
a
)
How To
Given a function
f
,
find the equation of a line tangent to the function at
x
=
a
.
Find the derivative of
f
(
x
)
at
x
=
a
using
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
.
Evaluate the function at
x
=
a
.
This is
f
(
a
)
.
Substitute
(
a
,
f
(
a
)
)
and
f
′
(
a
)
into
y
=
f
'
(
a
)
(
x
−
a
)
+
f
(
a
)
.
Write the equation of the tangent line in the form
y
=
m
x
+
b
.
Finding the equation of a line tangent to a function at a point
Find the equation of a line tangent to the curve
f
(
x
)
=
x
2
−
4
x
at
x
=
3.
Using:
f
'
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
Substitute
f
(
a
+
h
)
=
(
a
+
h
)
2
−
4
(
a
+
h
)
and
f
(
a
)
=
a
2
−
4
a
.
f
′
(
a
)
=
lim
h
→
0
(
a
+
h
)
(
a
+
h
)
−
4
(
a
+
h
)
−
(
a
2
−
4
a
)
h
=
lim
h
→
0
a
2
+
2
a
h
+
h
2
−
4
a
−
4
h
−
a
2
+
4
a
h
Remove parentheses
.
=
lim
h
→
0
a
2
+
2
a
h
+
h
2
−
4
a
−
4
h
−
a
2
+
4
a
h
Combine like terms
.
=
lim
h
→
0
2
a
h
+
h
2
−
4
h
h
=
lim
h
→
0
h
(
2
a
+
h
−
4
)
h
Factor out
h
.
=
2
a
+
0
−
4
f
′
(
a
)
=
2
a
−
4
Evaluate the limit
.
f
′
(
3
)
=
2
(
3
)
−
4
=
2
Equation of tangent line at
x
=
3:
y
=
f
'
(
a
)
(
x
−
a
)
+
f
(
a
)
y
=
f
'
(
3
)
(
x
−
3
)
+
f
(
3
)
y
=
2
(
x
−
3
)
+
(
−
3
)
y
=
2
x
−
9
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Source:
OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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