Ch. 11 – Limits and an
Introduction to Calculus
11.3 – The Tangent Line Problem
• WELCOME TO CALCULUS!!!
• Calculus is the study of rates of change of functions
• Recall: slope of a line determines the rate at which a line
rises or falls
• Today, we will use slope to study the rates of change of
curves
• Ex: At which point is the highest rate of change?
– Observe the slopes of each line…
– Draw in tangent lines to see slope
– The only positive slope is at A,
so the answer is A
B
C
Zoom in really, really close at 1 point.
It’s basically a line, right? That’s the
tangent line!
A
D
To find the slope of the tangent line, consider
secant lines that have almost the same slope
as the tangent line:
ff((xxhh)) ff((xx))
mmsecsec 
( x  hh)  x
(x+h, f(x+h))
As h approaches zero, we approach the true
slope of the tangent line:
mtan
f ( x  h)  f ( x )
 lim
h 0
h
(x+h, f(x+h))
(x+h, f(x+h))
(x+h, f(x+h))
(x, f(x))
(x+h, f(x+h))
x
x+h
x+h
x+h
x+h
x+h
• Ex: Find the slope of the graph of f(x) = x2 at the point (3, 9).
– Use the limit of the difference quotient from the last slide!
2
2
2
f (3  h)  f (3)
(3

h
)

3
9

6
h

h
9
lim
 lim
 lim
h 0
h 0
h 0
h
h
h
6h  h 2
 lim
h 0
h
h(6  h)
 lim
h 0
h
 lim(6  h)
h 0
6
1
f ( x) 
• Ex: Find a formula for the slope of the graph of
x
1
1

f ( x  h)  f ( x )
x  ( x  h)
x

h
x
lim
 lim
 lim
h 0
h 0 hx ( x  h)
h 0
h
h
Multiply by x(x+h)
h
 lim
h 0 hx( x  h)
1
 lim
h 0 x ( x  h )
1
 2
x
.
• The rate of change, or slope, of a function is called its
derivative. It is denoted by f’(x), which is read as “f prime of x”.
f ( x  h)  f ( x )
f '( x)  lim
h 0
h
• Ex: Find the derivative of f(x) = 2x2 – x .
2
2


2(
x

h
)

(
x

h
)

(2
x
 x)
f ( x  h)  f ( x )


f '( x)  lim
 lim
h 0
h 0
h
h
2 x 2  4 xh  2h2  x  h  2 x 2  x
 lim
h 0
h
 lim(4 x  2h  1)
h 0
 4 x 1
4 xh  h 2  2h
 lim
h 0
h
• Ex: Find the derivative of f ( x )  x and use it to write an
equation of the tangent line through (4, 2).
xh  x
f '( x)  lim
h 0
h
h
 lim
h 0 h( x  h 
x)
(Multiply by
conjugate of
numerator)
( x  h)  x
 lim
h 0 h( x  h 
x)
1
 lim
h 0
xh  x

1
x x
1

2 x
– Now evaluate at x=4 to find slope…
– Slope = ¼
– The tangent line has a slope of ¼ and passes through (4. 2), so write an
equation…
– I’ll use point slope form:
1
y  2  ( x  4)
4
1
y  2  x 1
4
1
y  x 1
4