2.1 – Rates of Change and Tangents to CurvesFunction Review

𝑓 (𝑥 )=2𝑥−7𝑓 (3 )=?

𝑓 (3 )=2 (3 )−7

𝑓 (3 )=6−7

𝑓 (3 )=−1

(3 ,−1 )

𝑓 (𝑥 )=2𝑥−7𝑓 (𝑎 )=?

𝑓 (𝑎 )=2 (𝑎 )−7

𝑓 (𝑎 )=2𝑎−7(𝑎 ,2𝑎−7 )

(𝑎 , 𝑓 (𝑎) )

𝑓 (𝑥 )=2𝑥−7𝑓 (3+𝑎)=?

𝑓 (3+𝑎)=2 (3+𝑎 )−7

𝑓 (3+𝑎)=6+2𝑎−7(3+𝑎 ,2𝑎−1 )

(3+𝑎 , 𝑓 (3+𝑎 ) )

2.1 – Rates of Change and Tangents to CurvesFunction Review

𝑓 (𝑥 )=𝑥2+2 𝑥−7𝑓 (𝑥+h )=?

𝑓 (𝑥+h )=(𝑥+h )2+2 (𝑥+h )−7

𝑓 (𝑥+h )=𝑥2+2 h𝑥 +h2+2 𝑥+2h−7

(𝑥+h , 𝑥2+2 h𝑥 +h2+2𝑥+2h−7   )

(𝑥+h , 𝑓 (𝑥+h ) )

2.1 – Rates of Change and Tangents to CurvesFunction Review

𝑓 (𝑥 )=𝑥3+3𝑥𝑓 (𝑥+h )=?

𝑓 (𝑥+h )=(𝑥+h )3+3 (𝑥+h )

𝑓 (𝑥+h )=𝑥3++3 𝑥2h+3 𝑥h2+h3+3 𝑥+3h(𝑥+h , 𝑥3++3 𝑥2h+3𝑥 h2+h3+3 𝑥+3h   )

(𝑥+h , 𝑓 (𝑥+h ) )

Pascal’s Triangle

2.1 – Rates of Change and Tangents to CurvesRate of change:

describes how one quantity changes in relation to another quantity changing. 

Examples:Rate: how distance changes as time changes

   

Rate: how area changes as time changes

/ /

Rate: how volume changes as time changes

/ /

Rate: how y changes as x changes𝑠𝑙𝑜𝑝𝑒 , ∆ 𝑦∆𝑥

2.1 – Rates of Change and Tangents to CurvesSecant Line: A line joining two points on a curve.

Secant line Secant line

Secant line

tangent line

2.1 – Rates of Change and Tangents to CurvesTangent Line: A line that touches a curve a one point.

tangent line

2.1 – Rates of Change and Tangents to CurvesRate of Change of a Secant Line or the Slope of a Secant Line

Secant line

𝑥1 𝑥2

∆ 𝑥=𝑥2−𝑥1h=𝑥2−𝑥1

(𝑥1 , 𝑓 (𝑥1 ))

(𝑥2 , 𝑓 (𝑥2 ))𝑓 (𝑥2)

𝑓 (𝑥1 )

𝑦= 𝑓 (𝑥 )

𝑠𝑙𝑜𝑝𝑒𝑜𝑓 𝑎𝑙𝑖𝑛𝑒 : 𝑚=∆ 𝑦∆ 𝑥

𝑚=𝑦2− 𝑦1𝑥2− 𝑥1

¿𝑓 (𝑥2 )− 𝑓 (𝑥1 )

𝑥2−𝑥1

𝑚=𝑓 (𝑥2 )− 𝑓 (𝑥1 )

∆ 𝑥¿𝑓 (𝑥2 )− 𝑓 (𝑥1 )

h

∆ 𝑦=𝑦2−𝑦1

𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑟𝑎𝑡𝑒𝑜𝑓 h𝑐 𝑎𝑛𝑔𝑒=𝑚=𝑓 (𝑥2 )− 𝑓 (𝑥1 )

𝑥2− 𝑥1

2.1 – Rates of Change and Tangents to Curves Average Rate of Change

Secant line

𝑥1 𝑥2

∆ 𝑥=𝑥2−𝑥1h=𝑥2−𝑥1

(𝑥1 , 𝑓 (𝑥1 ))

(𝑥2 , 𝑓 (𝑥2 ) )𝑓 (𝑥2)

𝑓 (𝑥1 )

𝑦= 𝑓 (𝑥 )

∆ 𝑦=𝑦2−𝑦1

𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑟𝑎𝑡𝑒𝑜𝑓 h𝑐 𝑎𝑛𝑔𝑒 𝑓𝑜𝑟 h𝑡 𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑛𝑑 h𝑡 𝑒𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 .𝑓 (𝑥 )=−2 𝑥2+4 [1,4 ]

𝑚=𝑓 (4 )− 𝑓 (1 )4−1

¿−28−23 ¿

−303 ¿−10

Example:

2.1 – Rates of Change and Tangents to Curves Average Rate of Change

Secant line

𝑥1 𝑥2

∆ 𝑥=𝑥2−𝑥1h=𝑥2−𝑥1

(𝑥1 , 𝑓 (𝑥1 ))

(𝑥2 , 𝑓 (𝑥2 ) )𝑓 (𝑥2)

𝑓 (𝑥1 )

𝑦= 𝑓 (𝑥 )

∆ 𝑦=𝑦2−𝑦1

𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑟𝑎𝑡𝑒𝑜𝑓 h𝑐 𝑎𝑛𝑔𝑒 𝑓𝑜𝑟 h𝑡 𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑛𝑑 h𝑡 𝑒𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 .𝑓 (𝑥 )=sec 𝑥 [0 , 𝜋3 ]

𝑚=𝑓 (𝜋3 )− 𝑓 (0 )

𝜋3 −0

¿

112

−1

𝜋3

¿2−1𝜋3

¿3𝜋

𝑓 (𝑥 )= 1cos 𝑥

¿1𝜋3

¿0.9549

Example:

¿

1

𝑐 𝑜𝑠 𝜋3

− 1cos (0)

𝜋3

2.1 – Rates of Change and Tangents to CurvesInstantaneous Rate of Change or the Slope of a Tangent Line

Secant line

Secant line

Secant line

tangent line

h h

hh=0

2.1 – Rates of Change and Tangents to CurvesInstantaneous Rate of Change or the Slope of a Tangent Line

Secant line

𝑥1 𝑥2

h

(𝑥1 , 𝑓 (𝑥1 ))

(𝑥2 , 𝑓 (𝑥2 ))𝑓 (𝑥2)

𝑓 (𝑥1 )

𝑦= 𝑓 (𝑥 )

3+h

h

3 3+2

h=2

𝑦= 𝑓 (𝑥 )

(3 , 𝑓 (3 ) )

(3+2 , 𝑓 (3+2 ) )

3 5

h=5−3=2

𝑦= 𝑓 (𝑥 )

(3 , 𝑓 (3 ) )

(5 , 𝑓 (5 ) )

(3 , 𝑓 (3 ) )

(3+h , 𝑓 (3+h ) )

3

𝑦= 𝑓 (𝑥 )

2.1 – Rates of Change and Tangents to CurvesInstantaneous Rate of Change or the Slope of a Tangent Line

tangent line

h=0

Slope formula requires two points.

A tangent line has one point on the curve.

Create a second point by selecting a small value of h.

𝑓 (𝑥 )=−2 𝑥2+4 (1,2 )𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑟𝑎𝑡𝑒𝑜𝑓 h𝑐 𝑎𝑛𝑔𝑒 𝑓𝑜𝑟 h𝑡 𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑡 h𝑡 𝑒𝑔𝑖𝑣𝑒𝑛𝑝𝑜𝑖𝑛𝑡 .

h=0.1

𝑚𝑡𝑎𝑛=𝑓 (1+0.1 )− 𝑓 (1 )

(1+0.1 )−1¿1.58−20.1

¿−0.420.1 ¿−4.2

𝑓 (𝑥 )=−2 𝑥2+4 (1,2 ) h=0.01

𝑚𝑡𝑎𝑛=𝑓 (1+0.01 )− 𝑓 (1 )

(1+0.01 )−1¿1.9598−20.01

¿−0.4020.01 ¿−4.02

2.1 – Rates of Change and Tangents to CurvesInstantaneous Rate of Change or the Slope of a Tangent Line

𝑓 (𝑥 )=−2 𝑥2+4 (1,2 ) h

𝑚𝑡𝑎𝑛=𝑓 (1+h )− 𝑓 (1 )

(1+h )−1=¿

−2 (1+h )2+4− (−2 (1 )2+4 )h

=¿

−4h−2h2

h=¿

𝑚𝑡𝑎𝑛=−4−2h

−2 (1+2h+h2 )+4− (−2 (1 )2+4 )h

=¿

−2−4h−2h2+4−2h

=¿

h (−4−2h )h

=¿

𝑚𝑡𝑎𝑛=−4−2hh=0.1

h=0.01

𝑚𝑡𝑎𝑛=−4−2(0.1)𝑚𝑡𝑎𝑛=−4.2

𝑚𝑡𝑎𝑛=−4−2(0.01)𝑚𝑡𝑎𝑛=−4.02

h=0.001𝑚𝑡𝑎𝑛=−4−2(0.001)

𝑚𝑡𝑎𝑛=−4.002

2.2 – Limit of a Function and Limit Laws

Find the requested limits from the graph of the given function.

lim𝑥→ 0

𝑓 (𝑥 )=¿¿

lim𝑥→ 1

𝑓 (𝑥 )=¿¿

lim𝑥→−1

𝑓 (𝑥 )=¿¿

lim𝑥→ 2

𝑓 (𝑥 )=¿¿

lim𝑥→−2

𝑓 (𝑥 )=¿¿

−10

0

3

3

Defn: Limit

lim𝑥→𝑐

𝑓 (𝑥 )=𝐿

As the variable x approaches a certain value, the variable y approaches a certain value.

2.2 – Limit of a Function and Limit Laws

Given the following graph of a function, find the requested limit.

lim𝑥→ 2

𝑔 (𝑥 )=¿¿

𝑔 (2 )=¿

4

6

lim𝑥→𝑐

𝑓 (𝑥 )=𝐿

(2,6 )

2.2 – Limit of a Function and Limit Laws

Given the following graph of a function, find the requested limits.

lim𝑥→ 0

𝑓 (𝑥 )=¿¿ 𝑓 (0 )=¿

lim𝑥→−1

𝑓 (𝑥 )=¿¿

lim𝑥→1

𝑓 (𝑥 )=¿¿

lim𝑥→2

𝑓 (𝑥 )=¿¿

2 2

𝑗𝑢𝑚𝑝

2

𝑗𝑢𝑚𝑝

lim𝑥→𝑐

𝑓 (𝑥 )=𝐿

(0,2 )

𝑓 (−1 )=¿2(−1,2 )

𝑓 (1 )=¿3(1,3 )

𝑓 (2 )=¿3(2,3 )

2.2 – Limit of a Function and Limit LawsFind the requested limits for the given function.

lim𝑥→ 0

𝑓 (𝑥 )=¿¿

lim𝑥→1

𝑓 (𝑥 )=¿¿

lim𝑥→−1

𝑓 (𝑥 )=¿¿

lim𝑥→2

𝑓 (𝑥 )=¿¿

lim𝑥→−2

𝑓 (𝑥 )=¿¿

(0 )2−1=¿−1

(1 )2−1=¿0

(−1 )2−1=¿0

3(2 )2−1=¿

(−2 )2−1=¿3

𝑓 (𝑥 )=𝑥2−1

2.2 – Limit of a Function and Limit LawsA rational function is the ratio of two polynomial functions

𝑓 (𝑥 )= 𝑥+52 𝑥−4

lim𝑥→ 3

𝑥+52𝑥−4=¿ ¿

3+52(3)−4

=¿82=¿ 4

Substitution Theorem:If f(x) is a polynomial function or a rational function, then or .If f(x) is a rational function, then the denominator cannot equal zero.

2.2 – Limit of a Function and Limit Laws

2.2 – Limit of a Function and Limit Laws

𝑓 (𝑥 )=𝑘lim𝑥→𝑐

𝑘=𝑘

Constant Function

𝑓 (𝑥 )=𝑥lim𝑥→𝑐

𝑥=𝑐

Identity Function

Additional Limit Rules

2.2 – Limit of a Function and Limit Laws

2.2 – Limit of a Function and Limit Laws

lim𝑥→ 3

14−𝑥2

=¿¿

Find the following limits:

− 15

lim𝑥→ 5

𝑥𝑥2− 𝑥

=¿¿14

lim𝑥→−4

𝑥−6𝑥2−36

=¿¿ 12

lim𝑥→ 6

𝑥+2𝑥2+𝑥−10

=¿¿ 14

2.2 – Limit of a Function and Limit LawsFind the following limits:

lim𝑥→ 0

𝑥𝑥2−𝑥

=¿¿ −1

lim𝑥→ 6

𝑥−6𝑥2−36

=¿¿

− 17lim𝑥→−2

𝑥+2𝑥2−3 𝑥−10

=¿¿

112

lim𝑥→ 0

𝑥𝑥 (𝑥−1)=¿¿

lim𝑥→ 6

𝑥−6(𝑥+6 ) (𝑥−6 )

=¿¿

lim𝑥→−2

𝑥+2(𝑥+2 ) (𝑥−5 )

=¿¿

lim𝑥→ 0

1(𝑥−1)=¿¿

lim𝑥→ 6

1(𝑥+6 )

=¿¿

lim𝑥→−2

1(𝑥−5 )

=¿¿