University of British Columbia

### Question Description

Need help with my Calculus question - Iβm studying for my class.

There are 9 questions to be answered. Please Answer the attached questions. Show all work.

### Unformatted Attachment Preview

## Final Answer

The solutions are ready, please ask if something is unclear.

1,2: Mean Value Theorem requires π(π₯ ) be continuous on [π, π] and differentiable on (π, π).

1. π (π₯ ) = βπ₯ 2 + 1, π₯ β [β1,1].

This function is continuous on [β1,1] but not differentiable on entire (β1,1) .

Indeed, π (π₯ ) = |π₯ | + 1 is not differentiable at π = π:

π(0 + β) β π(0) |β|

=

β

β

+

β

which tends to 1 as β β 0 but tends to β1 as β β 0 .

[note that the statement of MVT does not hold here as

π (1)βπ (β1)

1β(β1)

=

2β2

2

= 0 but never

π β² (π₯ ) = 0]

π₯ 3 + 3, π₯ < 1

2. π (π₯ ) = { 2

.

π₯ + 1, π₯ β₯ 1

This function is not continuous at π = π because

π (π₯ ) β 13 + 3 = 4, π₯ β 1

but

π (1) = 12 + 1 = 2 β 4.

[note that the statement of MVT does hold here as

π (1)βπ (β1)

1β(β1)

=

2β2

2

= 0 and π β² (0) = 0]

3. π (π₯ ) = 2 ln (π₯ + 3) , π₯ β [β2,0].

This elementary function is defined, continuous and differentiable on ( β3, β) so it is

continuous on [β2,0] and differentiable on (β2,0).

2

π (π)βπ (π)

2 ln 3β2 ln 1

To find π, determine π β² (π ) = π+3 and

=

= ln 3. Solve for π:

πβπ

2

2

π β π π₯π§ π

= ln 3 , 2 = π ln 3 + 3 ln 3 , π =

β β1.18 β [β2,0].

π+3

π₯π§ π

4. π (π₯ ) =

1

β4βπ₯2

. It is defined where 4 β π₯ 2 > 0, i.e. π₯ β (β2,2) .

It is increasing for π₯ > 0, π β (π, π) because π₯ 2 increases, 4 β π₯ 2 decreases, β4 β π₯ 2

decreases and

1

β4βπ₯2

increases.

In turn, it is decreasing for π₯ < 0, π β ( βπ, π) because π₯ 2 decreases, 4 β π₯ 2 increases,

β4 β π₯ 2 increases and

1

β4βπ₯2

decreases.

[We can also prove this using derivative:

1 β²

3

1

2 )β 2 )

)

((

π π₯ = 4βπ₯

= β (4 β π₯ 2 )β2 β (β2π₯ ) =

2

β²(

which is positive for π₯ > 0 and negative for π₯ < 0.]

The graph:

π₯

3,

(4 β π₯ 2 )2

5. π (π₯ ) = (π₯ β 2)(π₯ β 3)2 .

Find the derivative: π β² (π₯) =

= (π₯ β 2) β² (π₯ β 3)2 + (π₯ β 2)((π₯ β 3)2 )β² =

= (π₯ β 3) 2 + 2(π₯ β 2)(π₯ β 3) =

= (π₯ β 3)(π₯ β 3 + 2π₯ β 4) =

= (π₯ β 3)(3π₯ β 7)...