Calculus.docx

Example cosh 2𝑥−1

Prove that tanh2 𝑥 = cosh 2𝑥+1 Solution tanh2 𝑥 sinh 𝑥 2 tanh 𝑥 = ( ) cosh 𝑥 2

sinh2 𝑥

= cosh2 𝑥 1 sinh2 𝑥 = (𝑒 𝑥 − 𝑒 −𝑥 )2 2 cosh2 𝑥 =

1 𝑥 (𝑒 + 𝑒 −𝑥 )2 2

1 𝑥 (𝑒 − 𝑒 −𝑥 )2 cosh 2𝑥 − 1 2 = = 1 𝑥 (𝑒 + 𝑒 −𝑥 )2 cosh 2𝑥 + 1 2 1 2𝑥 (𝑒 + 𝑒 −2𝑥 − 2) 2 = 1 2𝑥 (𝑒 + 𝑒 −2𝑥 + 2) 2 𝑒 2𝑥 𝑒 −2𝑥 + 2 −1 = 22𝑥 𝑒 𝑒 −2𝑥 + 2 2 +1 1 2𝑥 (𝑒 + 𝑒 −2𝑥 ) − 1 2 1 2𝑥 (𝑒 + 𝑒 −2𝑥 ) + 1 2 =

cosh 2𝑥 − 1 cosh 2𝑥 + 1

Hence proved Example 21.1.3 Prove thatcosh(𝐴 + 𝐵) = cosh 𝐴 cosh 𝐵 + sinh 𝐴 sinh 𝐵 Solution From the left-hand side

𝑒 𝐴 + 𝑒 −𝐴 𝑒 𝐵 + 𝑒 −𝐵 𝑒 𝐴 − 𝑒 −𝐴 𝑒 𝐵 − 𝑒 −𝐵 cosh 𝐴 cosh 𝐵 + sinh 𝐴 sinh 𝐵 = ( )( )+( )( ) 2 2 2 2 𝑒 𝐴+𝐵 + 𝑒 𝐴−𝐵 + 𝑒 𝐵−𝐴 + 𝑒 −𝐴−𝐵 𝑒 𝐴+𝐵 − 𝑒 𝐴−𝐵 − 𝑒 𝐵−𝐴 + 𝑒 −𝐴−𝐵 =( )+( ) 4 4 =

2𝑒 𝐴+𝐵 + 2𝑒 −𝐴−𝐵 4

=

2(𝑒 𝐴+𝐵 + 𝑒 −𝐴−𝐵 ) 4

(𝑒 𝐴+𝐵 + 𝑒 −(𝐴+𝐵) ) = 2 = cosh(𝐴 + 𝐵). Hence cosh(𝐴 + 𝐵) = cosh 𝐴 cosh 𝐵 + sinh 𝐴 sinh 𝐵

INVERSE HYPERBOLIC FUNCTIONS 𝑦 = cosh−1 𝑥 means 𝑥 = cosh 𝑦 , 𝑦 ≥ 0, 𝑥 ≥ 1

GRAPHS OF HYPERBOLIC FUNCTIONS 𝑦 = 𝑐𝑜𝑠ℎ𝑥 y

1

𝑦 = 2 𝑒𝑥

𝑦 = 𝑠𝑖𝑛ℎ𝑥 0

y 𝑦 = 𝑐𝑜𝑡ℎ𝑥 𝑦=1

x 𝑦 = −1

𝑦 = 𝑐𝑜𝑡ℎ𝑥

y

𝑦 = 𝑐𝑜𝑠ℎ𝑥 𝑦=1 𝑦 = 𝑠𝑒𝑐ℎ𝑥 0

y

x

𝑦 = 𝑠𝑖𝑛ℎ𝑥

x

𝑦 = 𝑐𝑠𝑐ℎ𝑥

ELEMENTARY DIFFERENTIAL CALCULUS The Concept of a Limit Consider the functions 𝑓(𝑥) =

𝑥2 − 4 𝑥−2

The function is undefined at 𝑥 = 2 Let us examine the behavior of the function at 𝑥 = 2

𝑥

𝑓(𝑥) =

𝑥 2 −4 𝑥−2

𝑥

𝑓(𝑥) =

1.9

3.9

2.1

4.1

1.99

3.99

2.01

4.01

1.999

3.999

2.001

4.001

1.9999

3.9999

2.0001

4.0001

𝑥 2 −4 𝑥−2

Notice that as you move down the first column of the table, the 𝑥 −values get closer to 2 but are less than 2. We use the notation 𝑥 → 2− to indicate 𝑥 approaches 2 from the left. 𝑥 → 2+ to indicate 𝑥 approaches 2 from the right. We call lim 𝑓(𝑥) and 𝑙𝑖𝑚𝑓(𝑥) one sided limit. 𝑥 → 2− 𝑥 → 2− Theorem A limit exists if and only if both corresponding one sided limits exists and are equal. That is lim 𝑓(𝑥) = 𝐿 for some number 𝐿if and only if 𝑥→𝑎 lim 𝑓(𝑥) = 𝑙𝑖𝑚𝑓(𝑥) 𝑥 → a− 𝑥 → a+

Example Determine if the limit exists for the following a) lim

𝑥 2 −4

𝑥→2 𝑥−2 𝑥

b) lim |𝑥| 𝑥→0

Solution a) lim−

𝑥 2 −4 𝑥−2

𝑥→2

= lim−

(𝑥−2)(𝑥+2) 𝑥−2

𝑥→2

lim (𝑥 + 2) = 4

𝑥→2−

and lim+

𝑥 2 −4

𝑥→2

𝑥−2

= lim+

(𝑥−2)(𝑥+2)

𝑥→2

𝑥−2

lim (𝑥 + 2) = 4

𝑥→2−

∴ lim− = 4 = lim+ 𝑥→2

𝑥→2

Hence the limit exists.

𝑥

b) lim |𝑥| 𝑥→0

Solution 𝑥, |𝑥| = { −𝑥,

𝑥≥0 𝑥<0

𝑥 =1 𝑥→0 𝑥→0 𝑥 𝑥 lim− 𝑓(𝑥) = lim− = −1 𝑥→0 𝑥→0 −𝑥 lim+ 𝑓(𝑥) = lim+

Since lim+ 𝑓(𝑥) ≠ lim− 𝑓(𝑥) the limit does not exist. 𝑥→0

𝑥→0

y 1 𝑓(𝑥) 0←𝑥

𝑥→0 𝑓(𝑥)

-1 By definition 𝑥

𝑥

𝑓(𝑥) = 𝑥 = 1 and 𝑓(𝑥) = −𝑥 = −1

Theorem 1.2 sin 𝑥 =1 𝑥→0 𝑥 lim

Example Evaluate the following a) lim

𝑥→0

cos2 𝑥

b) lim

𝑥

𝑥→0

Solution a) lim

𝑥→0

b) lim

𝑥→0

cos2 𝑥 𝑥

tan3 𝑥 𝑥

= 2 lim

cos2 𝑥 2𝑥

𝑥→0

tan3 𝑥

= lim3 ( 𝑥→0

3𝑥

=2

)=3

tan3 𝑥 𝑥

COMPUTATION OF LIMITS For any constant 𝑐 and any real number 𝑎, lim 𝑐 = 𝑐

(1)

𝑥→𝑎

For any real number 𝑎, lim 𝑥 = 𝑎

(2)

𝑥→𝑎

Theorem 2.1 lim 𝑓( 𝑥) and lim 𝑔( 𝑥) both exists, and let 𝑐 be any constant. The

Suppose that

𝑥→𝑎

𝑥→𝑎

following then apply; i)

lim [𝑐. 𝑓(𝑥)] = 𝑐 lim 𝑓(𝑥)

ii)

lim [𝑓(𝑥) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥)

iii)

lim [𝑓(𝑥). 𝑔(𝑥)] = [lim 𝑓(𝑥)[ lim 𝑔(𝑥)]

𝑥→𝑎

𝑥→𝑎

𝑥→𝑎

𝑥→𝑎

𝑥→𝑎

𝑓(𝑥)

lim 𝑔(𝑥) =

iv)

𝑥→𝑎

𝑥→𝑎 lim 𝑓(𝑥)

𝑥→𝑎

lim 𝑔(𝑥)

𝑥→𝑎

𝑥→𝑎

𝑥→𝑎

if lim 𝑔(𝑥) ≠ 0 𝑥→𝑎

Example Apply the rule of limits to evaluate a) lim(3𝑥 2 − 5𝑥 + 4) 𝑥→2

b) lim

𝑥→3

c) lim

𝑥 3 −5𝑥+4 𝑥 2 −2 𝑥 2 −1

𝑥→1 1−𝑥 √𝑥+2−√2 𝑥 𝑥→0

d) lim

Solution a) lim(3𝑥 2 − 5𝑥 + 4) = lim3𝑥 2 − lim5𝑥 + lim4 𝑥→2

= 12 − 10 + 4 =6

𝑥→2

𝑥→2

𝑥→2

b) lim

𝑥 3 −5𝑥+4

=

𝑥 2 −2

𝑥→3

c) lim

𝑥 2 −1

𝑥→1 1−𝑥

(3)3 −5(3)+4

= 𝑙𝑖𝑚

(3)2 −2 (𝑥−1)(𝑥+1) −(𝑥−1)

𝑥→1

=

16 7

= −2

√𝑥+2−√2 𝑥 𝑥→0

d) lim

lim

(√𝑥+2−√2)(√𝑥+2+√2) 𝑥(√𝑥+2+√2)

𝑥→0

𝑥+2−2

lim 𝑥(√𝑥+2+√2)

𝑥→0

𝑥

lim

𝑥→0 𝑥(√𝑥+2+√2)

1

lim

𝑥→0 (√𝑥

+ 2 + √2)

1 √2 + √2

=

1 2√2

Theorem 2.2 For any polynomial 𝑝(𝑥) and any real number 𝑎, lim 𝑝(𝑥) = 𝑝(𝑎)

𝑥→𝑎

Theorem 2.3 Suppose that lim 𝑓(𝑥) = 𝐿 and 𝑛 is any positive integer, Then 𝑥→𝑎

𝑛

𝑛

lim √𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥) = √𝐿 𝑥→𝑎 𝑥→𝑎 Where for 𝑛 even, we assume that 𝐿 > 0

Theorem 2.4 For any real number 𝑎, we have; i) ii)

lim sin 𝑥 = sin 𝑎

𝑥→𝑎

lim cos 𝑥 = cos 𝑎

𝑥→𝑎

lim 𝑒 𝑥 = 𝑒 𝑎

iii)

𝑥→𝑎

lim ln 𝑥 = ln 𝑎 for 𝑎 > 0

iv)

𝑥→𝑎

lim sin−1 𝑥 = sin−1 𝑎 for −1 < 𝑎 < 1

v)

𝑥→𝑎

lim cos−1 𝑥 = cos−1 𝑎 for −1 < 𝑎 < 1

vi)

𝑥→𝑎

lim tan−1 𝑥 = tan−1 𝑎 for −∞ < 𝑎 < ∞

vii)

𝑥→𝑎

and If P is a polynomial and lim 𝑓(𝑥) = 𝐿, then 𝑥→𝑎

lim 𝑓(𝑝(𝑥)) = 𝐿

𝑥→𝑎

Example 5

a) Evaluate lim √3𝑥 2 − 2𝑥 𝑥→2

b) Evaluate lim sin−1 (

𝑥+1

𝑥→0

2

)

c) Evaluate lim(𝑥𝑐𝑜𝑡𝑥) 𝑥→0

Solution 5

5

a) lim √3𝑥 2 − 2𝑥 = 5√lim3𝑥 2 − 2𝑥 = √8 𝑥→2 𝑥→2 b) lim sin−1 ( 𝑥→0

𝑥+1 2

1

) = sin−1 (2) =

𝜋 6

c) lim(𝑥𝑐𝑜𝑡𝑥) = (lim 𝑥) (lim𝑐𝑜𝑡𝑥) = 0.0 ? 𝑥→0

𝑥→0

𝑥→0

cos 𝑥

𝑥

∴ lim(𝑥𝑐𝑜𝑡𝑥) = lim (𝑥. sin 𝑥 ) = lim (sin 𝑥) (cos 𝑥) 𝑥→0

𝑥→0

𝑥 (lim ) (lim cos 𝑥) sin 𝑥 𝑥→0 𝑥→0

1 sin 𝑥 ) (lim cos 𝑥) = 1 lim ( 𝑥→0 𝑥→0 𝑥

𝑥→0

Theorem 2.5 SQUEEZ THEOREM Suppose that 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) For all 𝑥 in some interval (𝑐, 𝑑), except possibly at the point 𝑎 ∈ (𝑐, 𝑑) and that lim 𝑓(𝑥) = lim ℎ(𝑥) = 𝐿

𝑥→𝑎

𝑥→𝑎

For some number 𝐿. Then it follows that lim 𝑔(𝑥) = 𝐿 also.

𝑥→𝑎

Example 1

Evaluate 𝑡ℎ𝑒 lim [𝑥 2 cos (𝑥)] 𝑥→0

Solution 1 lim [𝑥 2 cos ( )] 𝑥→0 𝑥 1 (−1 ≤ cos ( ) ≤ 1) 𝑥 2 𝑥 1 −𝑥 2 ≤ 𝑥 2 cos ( ) ≤ 𝑥 2 𝑥 𝑥≠0 lim−𝑥 2 = 0 = lim𝑥 2

𝑥→0

𝑥→0

∴ from the squeeze theorem, it follows that 1 lim𝑥 2 cos ( ) = 0 𝑥→0 𝑥

A LIMIT FOR A PIECEWISE FUNCTION Example Evaluate lim𝑓(𝑥), where 𝑓(𝑥) is defined by 𝑥→0

𝑥 2 + 2𝑐𝑜𝑠𝑥 + 1, 𝑓𝑜𝑟 𝑥 < 0, 𝑓(𝑥) = 𝑓(𝑥) = { 𝑒 𝑥 − 4,

𝑓𝑜𝑟 𝑥 ≥ 0

Solution lim 𝑓(𝑥) = lim− (𝑥 2 + 2𝑐𝑜𝑠𝑥 + 1) = 3

𝑥→0−

𝑥→0

and also lim (𝑒 𝑥 − 4) = 𝑒 0 − 4 = 1 − 4 = −3

𝑥→0−

Thus lim𝑓(𝑥) does not exist 𝑥→0

LIMITS INVOLVING INFINITY; ASYMPTOTES Example Examine lim

1

𝑥→0 𝑥

solution

𝑓(𝑥)

0 𝑓(𝑥)

lim+

𝑥→0

lim−

𝑥→0

1 =∞ 𝑥 1 𝑥

= −∞ the limit does not exist 1

The graph of 𝑦 = 𝑥 approaches the vertical line 𝑥 = 0, as 𝑥 → 0 as shown above When this occurs, we say that the line 𝑥 = 0.

Example Evaluate lim

1

𝑥→0 𝑥 2

𝑓(𝑥)

𝑓(𝑥)

solution 0

1 =∞ 𝑥→0 𝑥 2 1 lim− 2 = −∞ 𝑥→0 𝑥 The limit does not exist. lim+

LIMITS AT INFINTY 1

1

Returning to 𝑓(𝑥) = 𝑥 , we can see that as 𝑥 → ∞, 𝑥 → 0. In view of this, we write 1 =0 𝑥→∞ 𝑥 lim

1 =0 𝑥→−∞ 𝑥 lim

The graph appears to approach the horizontal line 𝑦 = 0, as 𝑥 → ∞ and as 𝑥 → −∞. In this case, we call 𝑦 = 0 a horizontal asymptote. A LIMIT OF A QUOTIENT THAT IS NOT THE QUOTIENT OF A LIMIT Example Evaluate lim

5𝑥−7

𝑥→∞ 4𝑥+3

Solution lim

𝑥→∞

lim (5𝑥−7)

5𝑥−7

∞

= 𝑥→∞ =∞ ? 4𝑥+3 lim (4𝑥+3) 𝑥→∞

NOTE: i) ii) iii) iv)

∞

indeterminate form

∞ ∞

0 ∞. ∞ 0.0

Rule of thumbs ∞

When faced with the indeterminate form ∞ in calculating limits of a rational function, divide numerator and denominator by the highest power of x appearing in the denominator.

Here we have 5𝑥 7 7 −𝑥 5−𝑥 5 5𝑥 − 7 𝑥 lim = = lim = 4𝑥 3 𝑥→∞ 3 4 𝑥→∞ 4𝑥 + 3 4+𝑥 𝑥 +𝑥

Example Evaluate lim

4𝑥 3 +5

𝑥→∞ −6𝑥 2 −7𝑥

Solution 4𝑥 3 5 + 2 4𝑥 3 + 5 𝑥2 𝑥 lim = lim 𝑥→∞ −6𝑥 2 − 7𝑥 𝑥→∞ 6𝑥 2 7𝑥 − 2 − 2 𝑥 𝑥 5 2 𝑥 lim = −∞ 7 𝑥→∞ −6 − 𝑥 4𝑥 +

Two LIMITS OF AN INVERSE TRIGONOMETRIC FUNCTION Evaluate lim tan−1 𝑥 and lim tan−1 𝑥 𝑥→∞

𝑥→∞

y

y = tan−1 𝑥

𝜋

𝜋

−2

lim tan−1 𝑥 =

𝑥→∞

2

𝜋 2

lim tan−1 𝑥 = −

𝑥→−∞

𝜋 2

𝜋 2

𝑦 = tan−1 𝑥

𝜋

−2

Example Evaluate lim (√𝑛 + 1 − √𝑛) 𝑥→∞

Solution (√𝑛+1−√𝑛)

lim (√𝑛 + 1 − √𝑛) (

𝑥→∞

lim

𝑛 + 1 + √𝑛 √𝑛 + 1 − √𝑛 √𝑛 + 1 − 𝑛 √𝑛 + 1 + √𝑛

𝑥→∞

lim

√𝑛+1−√𝑛)

1

𝑥→∞ √𝑛

+ 1 + √𝑛 1

𝑛+1 √𝑛 √ 𝑛 + √𝑛 lim

𝑥→∞

lim

𝑥→∞

1 1 √𝑛 √ ( 1 + 𝑛 + 1) 𝑛 √ 1 √𝑛 1 √𝑛 √ ( 1 + 𝑛 + 1) √𝑛

=

0 =0 2

CONTINUITY Definition A function f is continuous at 𝑥 = 𝑎 when 𝑓(𝑥) is defined lim 𝑓(𝑥) exists

i) ii)

𝑥→𝑎

lim 𝑓(𝑥) = 𝑓(𝑎)

iii)

𝑥→𝑎

Otherwise,𝑓 is said to be discontinuous at 𝑥 = 𝑎 As an example of a continuous function, let us investigate 𝑓(𝑥) = 𝑥 2 At some fixed value 𝑥 = 𝑐. certainly 𝑓(𝑐) = 𝑐 2 at 𝑥 = 𝑐. Furthermore, the difference 𝑓(𝑥) − 𝑓(𝑐) = 𝑥 2 − 𝑐 2 = (𝑥 − 𝑐)(𝑥 + 𝑐) Approaches zero as a limit when 𝑥 − 𝑐 approaches zero, since as when 𝑥 ⟶ 𝑐, We have lim[𝑓(𝑥) − 𝑓(𝑐)] = lim(𝑥 − 𝑐)lim(𝑥 + 𝑐) 0.2𝑐 = 0 ∴ 𝑓(𝑥) = 𝑥 2 is continuous at any point where 𝑥 = 𝑐

Example 3.1 Finding where a rational function is continuous Determine where 𝑓(𝑥) =

𝑥 2 +2𝑥−3 𝑥−1

is continuous

Solution 𝑓(𝑥) =

𝑥 2 +2𝑥−3 𝑥−1

=

(𝑥−1)(𝑥+3) 𝑥−1

=𝑥+3

for 𝑥 ≠ 1 this says that the graph of 𝑓 is a straight line but with a hole in it at 𝑥 = 1 ∴ 𝑓 is discontinuous at 𝑥 = 1 but continuous everywhere

{𝐷: 𝑥 ∈ ℝ|𝑥 ≠ 1} y

𝑦=

𝑥 2 +2𝑥−3 𝑥−1

4

5

x

Example Removing a discontinuity Let us redefine example 3.1 to make it continuous everywhere

𝑥 2 + 2𝑥 − 3 , 𝑖𝑓 𝑥 ≠ 1 𝑔(𝑥) = { 𝑥 − 1 𝑎, 𝑖𝑓 𝑥=1

lim𝑔(𝑥) = lim

𝑥→1

𝑥2 + 2𝑥 − 3

𝑥→1

𝑥−1

=

(𝑥 − 1)(𝑥 + 3) 𝑥→1 𝑥−1 lim

lim(𝑥 + 3) = 4

𝑥→1

Now if we choose 𝑎 = 4,then lim𝑔(𝑥) = 4 = 𝑔(1) and so g is continuous at 𝑥 = 1.

𝑥→1

NOTE: When we can remove a discontinuity by redefining the function at that point, we call the discontinuity a removable discontinuity. Example The function f defined by 𝑓(𝑥) =

sin 𝑥 𝑥

for 𝑥 ≠ 0

has a removable discontinuity at 𝑥 = 0 , the formula is not valid when 𝑥 = 0 but lim

𝑥→0

sin 𝑥 𝑥

=1

sin 𝑥 𝑓(𝑥) = { 𝑥 , 1,

𝑤ℎ𝑒𝑛 𝑥 ≠ 0 𝑤ℎ𝑒𝑛 𝑥 = 0

by so defining the value of f at 0, we have made f continuous there, since we now have 𝑓(𝑥) = 𝑙𝑖𝑚𝑓(𝑥) as 𝑥 approaches zero. Non-removable discontinuity The discontinuity at 𝑥 = 0 in the foregoing example could be removed because f had a limit, namely 1, as 𝑥 approaches zero. On the other hand, if 𝑔(𝑥) =

|𝑥| 𝑥

for 𝑥 ≠ 0

Then there is a discontinuity at 0, that can be removed. 𝑥, 𝑖𝑓 𝑥 ≥ 0 |𝑥| = { −𝑥, 𝑥<0 𝑥 =1 𝑥→0 𝑥 𝑥 lim− = 1 𝑥→0 𝑥 lim+

∴ the limit does not exist. If we assign the value 0 to g(0), the resulting function is the signum function ( or sign function) of 𝑥. +1 sign𝑥 = { 0 −1

𝑖𝑓 𝑥 > 0 𝑖𝑓 𝑥 = 0 𝑖𝑓 𝑥 < 0

the signum function is defined for all 𝑥, but it is not continuous at 𝑥 = 0. It is continuous for 𝑥 ≠ 0 Example Non-removable discontinuity 𝑓(𝑥) =

1 𝑥2

Solution We should observe from the diagram 1 =∞ 𝑥→0 𝑥 2 lim

∴ the limit does not exist .

-3

3

Hence 𝑓 is discontinuous at 𝑥 = 0 and non- removable. Theorem All polynomials are continuous everywhere. Additionally, 𝑠𝑖𝑛𝑥, 𝑐𝑜𝑠𝑥, tan−1 𝑥 𝑎𝑛𝑑 𝑒 𝑥 are 𝑛 continuous everywhere. √𝑥 is continuous for all 𝑥, when n is odd and for 𝑥 ≥ 0 when n is even. We also have that ln 𝑥 is continuous for 𝑥 > 0 and sin−1 𝑥 and cos −1 𝑥 are continuous for −1 < 𝑥 < 1 Theorem Suppose that 𝑓 𝑎𝑛𝑑 𝑔 are continuous at 𝑥 = 𝑎. Then all of the following are true: i) (𝑓 ± 𝑔) is continuous at x=a ii) (𝑓. 𝑔) is continuous at x=a and 𝑓

iii) (𝑔) is continuous at 𝑥 = 𝑎 if 𝑔(𝑎) ≠ 0 Theorem Suppose that lim 𝑔(𝑥) = 𝐿 and 𝑓 is continuous at 𝐿 then, 𝑥→𝑎

lim 𝑓(𝑔(𝑥)) = 𝑓(𝑙𝑖𝑚𝑔(𝑥)) = 𝑓(𝐿) 𝑥→𝑎

𝑥→𝑎

Corollary 3.1 Suppose that 𝑔 is continuous at a and 𝑓 is continuous at 𝑔(𝑎). Then, the composition 𝑓𝑜𝑔 is continuous at a. Example Continuity for a composite function Determine where ℎ(𝑥) = cos(𝑥 2 − 5𝑥 + 2) is continuous. Solution Note that ℎ(𝑥) = 𝑓(𝑔(𝑥)) Where 𝑔(𝑥) = 𝑥 2 − 5𝑥 + 2 and 𝑓(𝑥) = 𝑐𝑜𝑠𝑥 Since both 𝑓 𝑎𝑛𝑑 𝑔 are continuous for all x, h is also continuous for all x by corollary 3.1 Definition 3.2 If f is continuous at every point on an open interval (𝑎, 𝑏), we say that f is continuous on the closed interval [𝑎, 𝑏], if f is continuous on the open interval (𝑎, 𝑏) and lim 𝑓(𝑥) = 𝑓(𝑎) and lim−𝑓(𝑥) = 𝑓(𝑏).

𝑥→𝑎+

𝑥→𝑏

Finally, if 𝑓 is continuous on all of 𝑔(−∞, ∞), we simply say that 𝑓 is continuous. Example Determine the interval where 𝑓 is continuous for, a) 𝑓(𝑥) = √4 − 𝑥 2 b) 𝑓(𝑥) = ln(𝑥 − 3) solution a) 𝑓(𝑥) = √4 − 𝑥 2 √4 − 𝑥 2 ≥ 0 2

(√4 − 𝑥 2 ) ≥ (0)2 4 − 𝑥2 ≥ 0 (−1) − 𝑥 2 ≥ −4(−1) 𝑥2 ≤ 4 √𝑥 2 ≤ ±√4 |𝑥| ≤ ±2

recall that |𝑥| = √𝑥 2

−2 ≤ 𝑥 ≤ 2

b)

𝑓(𝑥) = ln(𝑥 − 3)

solution 𝑥−3>0 ⟹𝑥 >3 ∴ 𝑓 is continuous on the interval (3, ∞).

Theorem The intermediate value theorem Suppose that 𝑓 is continuous on the closed interval [𝑎, 𝑏] and w is any number between 𝑓(𝑎) and 𝑓(𝑏). Then there is a number 𝑐 ∈ [𝑎, 𝑏] for which 𝑓(𝑐) = 𝑤.

CORRALARY Suppose that 𝑓 is continuous on [𝑎, 𝑏] and 𝑓(𝑎) and 𝑓(𝑏) have opposite signs [𝑖. 𝑒 𝑓(𝑎). 𝑓(𝑏) < 0]. Then there is at least one number 𝑐 ∈ (𝑎, 𝑏) for which 𝑓(𝑐) = 0 (recall that c is then a zero of f) Note: Lemma lim

𝜃→0

1−cos 𝜃 𝜃

=0

DIFFERENTIATION THE GENERAL CASE To find the slope of the tangent line to 𝑦 = 𝑓(𝑥) 𝑎𝑡 𝑥 = 𝑎, first pick two points on the curve, one point is the point of tangency (𝑎, 𝑓(𝑐)). Call the x-coordinate of the secant point 𝑥 = 𝑎 + ℎ, for some small number corresponding y-coordinates is 𝑓(𝑎 + ℎ).It is natural to think of h as being positive , as shown in figure (a), h can also be negative as shown in figure (b)

●

● 𝑎

𝑎+ℎ

𝑎+ℎ

𝑎

a)

b) secant line ℎ > 0

secant line ℎ < 0

the slope of the secant line through the point (𝑎, 𝑓(𝑎)) and (𝑎 + ℎ, 𝑓(𝑎 + ℎ) is given by

𝑀𝑠𝑒𝑐 =

𝑓(𝑎+ℎ)−𝑓(𝑎) (𝑎+ℎ)−𝑎

=

𝑓(𝑎+ℎ)−𝑓(𝑎) ℎ

1.1

Notice that the expression in 1.1 is called a difference quotient gives the slope of the secant line for any secant point we might choose (that is for any ℎ ≠ 0)

In order to obtain an improved approximation to the tangent line, we zoom in closer and closer towards the point of tangency. This makes the two points closer together to 0 as shown below. y

𝑄

●

● 𝑃 ● ● ● x

Notice that as the point Q approaches the point P(𝑖. 𝑒 , 𝑎𝑠 ℎ → 0), the secant line approaches the tangent line at P. Definition The slope 𝑚𝑡𝑎𝑛 of the tangent line to 𝑦 = 𝑓(𝑥) at 𝑥 = 𝑎 is given by

𝑓(𝑎 + ℎ) − 𝑓(𝑎) ℎ→0 ℎ

𝑓 ′ (𝑎) = lim

Provided the limit exists. If the limit exists, we say that 𝑓 is differentiable at 𝑥 = 𝑎. 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ ℎ→0

Or the derivative of 𝑓(𝑥) is the function 𝑓 ′ (𝑥) given by 𝑓 ′ (𝑥) = lim

provided

the limit exists. The process of computing a derivative is called differentiation. Further, f is differentiable at a point on an interval 𝐼 if it is differentiable at every point in 𝐼. This is also called the 1𝑠𝑡 principle of derivative. Example Use the definition to find the derivative function of the following. a) 𝑓(𝑥) = √𝑥 𝑎𝑛𝑑 𝑥 > 0 b) 𝑓(𝑥) = sin 𝑥

solution

𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ→0 ℎ

𝑓 ′ (𝑥) = lim

√𝑥+ℎ−√𝑥 ℎ ℎ→0

= lim = lim

(√𝑥 + ℎ − √𝑥)(√𝑥 + ℎ + √𝑥) ℎ(√𝑥 + ℎ + √𝑥)

ℎ→0

(𝑥+ℎ)−𝑥

= lim ℎ(√𝑥+ℎ+

√𝑥)

ℎ→0

1

= lim

ℎ→0 √𝑥

= =

+ ℎ + √𝑥

1 √𝑥 + 0 + √𝑥 1 2 √𝑥

b) 𝑓(𝑥) = sin 𝑥 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ→0 ℎ

𝑓 ′ (𝑥) = lim

sin(𝑥 + ℎ) − 𝑠𝑖𝑛𝑥 ℎ→0 ℎ

= lim

𝑠𝑖𝑛𝑥 𝑐𝑜𝑠ℎ + 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 − 𝑠𝑖𝑛𝑥 ℎ→0 ℎ

= lim

(𝑐𝑜𝑠𝑥 − 1) 𝑠𝑖𝑛ℎ + 𝑐𝑜𝑠𝑥 lim ℎ→0 ℎ→0 ℎ ℎ

= 𝑠𝑖𝑛𝑥 lim

= 𝑠𝑖𝑛𝑥(0) + 𝑐𝑜𝑠𝑥(1) = 𝑐𝑜𝑠𝑥

ALTERNATIVE DERIVATIVE NOTATIONS We have denoted the derivative function by 𝑓 ′ (𝑥). There are other Commonly used notations, each with advantages and disadvantages. One of the 𝑑𝑓

converts of calculus, Leibniz, used the notation 𝑑𝑥 (Leibniz notation) for the derivative. If we write 𝑦 = 𝑓(𝑥), the following are all alternatives for denoting the derivative

𝑓 ′ (𝑥) = 𝑦 ′ =

𝑑𝑦 𝑑𝑥

=

𝑑𝑓 𝑑𝑥

=

𝑑 𝑑𝑥

𝑓(𝑥)

𝑑

The expression 𝑑𝑥 is called a differential operator and tells you to take the derivative of whatever expression follows. THEOREM If 𝑓(𝑥) is differentiable at 𝑥 = 𝑎 ,then 𝑓(𝑥) is continuous at 𝑥 = 𝑎. Example Show that 𝑓(𝑥) = {

4, 𝑖𝑓 𝑥 < 2 2𝑥 𝑖𝑓, 𝑥 ≥ 2

is not differentiable at 𝑥 = 2

Solution

𝑓(𝑥+ℎ)−𝑓(𝑥)

𝑓 ′ (𝑥) = lim−

ℎ

ℎ→0

lim−

𝑓(𝑥+ℎ)−𝑓(𝑥)

= lim−

ℎ

ℎ→0

𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ

ℎ→0

2𝑥+2ℎ−2𝑥

ℎ→0

ℎ

ℎ→0

𝑓 ′ (𝑥) = lim+ lim+

𝑓(2+ℎ)−𝑓(2)

= lim+

ℎ→0

2(𝑥+ℎ)−2(𝑥)

ℎ→0

=2

2

∴ 𝑓 ′ (𝑥) ≠ 𝑓 ′ (𝑥) lim

ℎ→0−

lim

ℎ→0+

The function is not continuous at 𝑥 = 2 COMPUTATION OF DERIVATIVES For any constant c, 𝑑 𝑑𝑥

(𝑐) = 0

Similarly, 𝑑 𝑑𝑥

𝑥=1

Theorem (power rule) For any integer 𝑛 > 0, 𝑑 𝑑𝑥

𝑥 𝑛 = 𝑛𝑥 𝑛−1

Theorem (General power rule) For any real number 𝑟 ,

𝑑 𝑑𝑥

= lim−

𝑥 𝑟 = 𝑟𝑥 𝑟−1

ℎ

4−4 ℎ

=0

Example Find the derivatives of the following a)

1 𝑥 19 3

b) √𝑥 2 c)

𝑥𝜋

solution a)

1 𝑥 19 𝑑

( 𝑑𝑥

19 1 𝑑 ) = 𝑑𝑥 𝑥 −19 = −19𝑥 −19−1 = −19𝑥 −20 = 𝑥 −20 19 𝑥 2

3

b) √𝑥 2 = 𝑥 3 𝑑 3

2

𝑑

2

2

2

1

√𝑥 2 = 𝑑𝑥 𝑥 3 = 3 𝑥 3−1 = 3 𝑥 −3 = 𝑑𝑥 c)

𝑑 𝑑𝑥

2 1 3𝑥 3

=

2 3

3 √𝑥

𝑥 𝜋 = 𝜋𝑥 𝜋−1

GENERAL DERIVATIVE RULES THEOREM If 𝑓(𝑥) and 𝑔(𝑥) are differentiable at 𝑥 and 𝑐 is any constant, then (i) (ii) (iii)

𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑 𝑑𝑥

[𝑓(𝑥) + 𝑔(𝑥)] = 𝑓 ′ (𝑥) + 𝑔′ (𝑥) [𝑓(𝑥) − 𝑔(𝑥)] = 𝑓 ′ (𝑥) − 𝑔′ (𝑥) and [𝑐𝑓(𝑥)] = 𝑐𝑓 ′ (𝑥)

Example Find the derivative of a) 𝑓(𝑥) = 2𝑥 6 + 3√𝑥 b) 𝑓(𝑥) =

4𝑥 2 −3𝑥+2√𝑥 𝑥

Solution 𝑑

𝑑

a) 𝑓 ′ (𝑥) = 𝑑𝑥 (2𝑥 6 ) + 𝑑𝑥 (3√𝑥) 𝑑

𝑑

1

1

= 2 𝑑𝑥 (𝑥 6 ) + 3 𝑑𝑥 (√𝑥) = 2(6𝑥 5 ) + 3 (2) 𝑥 −2 = 12𝑥 5 + 2

3

√𝑥

b)