1.6 – Tangent Lines and Slopes

•Slope of Secant Line•Slope of Tangent Line•Equation of Tangent Line•Equation of Normal Line•Slope of Tangent = 0

3.1

Given the curve:

x

f(x)

x+h

f(x+h)

sec

f x h f xm

h

tan

h 0

f x h f xm lim

h

Given the equation 2y x 4x, 4, 0 a) Find the slope of a secant line through the given point

2 2

sec

x h 4 x 4xx

h

hm

2

sec

2 2x 2xh h 4x x 4xm

h

4h 2

sec

2xh h 4hm

h

sec

h 2m

h

x h 4

sec 2 hm x 4

Given the equation 2y x 4x, 4, 0 b) Find the SLOPE of a tangent line through the given point

From part a, sec 2 hm x 4

tanh 0

2x hm lim 2x 44

At (4, 0), 2x – 4 = 2(4) – 4 = 4

c) Find the EQUATION of a tangent line through the given point

From part b, we have the slope (4)….and we have the pt (4, 0)

y – 0 = 4(x – 4)

c) Find the value of x for which the slope of the tangent line is 0

2x – 4 = 0x = 2

Given the equation 3y x 2x, 1, 1 a) Find the slope of a secant line through the given point

3 3

sec

x h 2 x 2xx

h

hm

3 2 2 3 3

sec

x 3x h 3xh h 2x x 2xm

h

2h 2

sec

2 33x h 3xh h 2hm

h

2 2

sec

h 3x 3x 2m

h

h

h 2

sec23x hm 3xh 2

b) Find the SLOPE of a tangent line through the given point

From part a,

tanh

2 22

03x 3xhm l m 2i 3h x 2

At (1, -1),

c) Find the EQUATION of a tangent line through the given point

From part b, we have the slope (1)….and we have the pt (1, -1)

y + 1 = 1(x - 1)

c) Find the value of x for which the slope of the tangent line is 0

Given the equation 3y x 2x, 1, 1

2sec

23x hm 3xh 2

223x 2 3 1 2 1

2 2 2 23x 2 0 x x

3 3

Using the limit definition, find the first derivative of 3f x x

3 3

h 0

x xl

h

him

3 33

h

2

0

2x 3x h 3xh xlim

h

h

2 2 3

h 0

3x h 3xh hlim

h

0

2 2

h

h 3x 3xh hlim

h

2 2 2

h 03x 3xhi xl h 3m

Using the limit definition, find the first derivative of f x 1 x

h 0

1 1 xlim

h

x h

h 0

xlim

h

x h

x

x x

h x

h

h 0

xlim

h x h

h

x

x

h 0

hlim

h x h x

h 0

1 1lim

2 xx h x

Using the limit definition, find the first derivative of f x x 1

h 0

x h 1 x 1lim

h

h 0

x h x1 x 1 1 x 1lim

h h xx 1

h

1

h 0

1 x 1lim

h 1

h

x 1

x

x h

h 0

h

x hlim

h 1 x 1

h 0

1 1lim

2 x 11 xx h 1

When will the slope of the tangent to be equal to zero?y x 1

•The slope of the tangent line is the first derivative•We just computed the first derivative.

dy 1y' f ' x

dx 2 x 1

1

02 x 1

•The numerator of the fraction is never zero•Therefore, the slope of the tangent is never zero.

Find the equation of the line normal to f x x x 1 at 3, 6

2f x x x

2 2

h 0

x xlim

h

x h x h

2

h 0

2 2x 2xh h x h x xlim

h

h 0

h 2x h 1lim 2x 1

h

tanm 2x 1 2 3 1 5 normtan

1 1m

m 5

1y 6 x 3

5