Mathematics
University of British Columbia Mean Value Theorem & Parabola Branches Worksheet

University of British Columbia

Question Description

Need help with my Calculus question - I’m studying for my class.

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

University of British Columbia Mean Value Theorem & Parabola Branches Worksheet
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University of British Columbia Mean Value Theorem & Parabola Branches Worksheet
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University of British Columbia Mean Value Theorem & Parabola Branches Worksheet
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University of British Columbia Mean Value Theorem & Parabola Branches Worksheet
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University of British Columbia Mean Value Theorem & Parabola Branches Worksheet
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1,2: Mean Value Theorem requires 𝑓(π‘₯ ) be continuous on [π‘Ž, 𝑏] and differentiable on (π‘Ž, 𝑏).
1. 𝑓 (π‘₯ ) = √π‘₯ 2 + 1, π‘₯ ∈ [βˆ’1,1].
This function is continuous on [βˆ’1,1] but not differentiable on entire (βˆ’1,1) .
Indeed, 𝑓 (π‘₯ ) = |π‘₯ | + 1 is not differentiable at 𝒙 = 𝟎:
𝑓(0 + β„Ž) βˆ’ 𝑓(0) |β„Ž|
=
β„Ž
β„Ž
+
βˆ’
which tends to 1 as β„Ž β†’ 0 but tends to βˆ’1 as β„Ž β†’ 0 .

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

𝑓 (1)βˆ’π‘“ (βˆ’1)
1βˆ’(βˆ’1)

=

2βˆ’2
2

= 0 but never

𝑓 β€² (π‘₯ ) = 0]

π‘₯ 3 + 3, π‘₯ < 1
2. 𝑓 (π‘₯ ) = { 2
.
π‘₯ + 1, π‘₯ β‰₯ 1
This function is not continuous at 𝒙 = 𝟏 because
𝑓 (π‘₯ ) β†’ 13 + 3 = 4, π‘₯ β†’ 1
but
𝑓 (1) = 12 + 1 = 2 β‰  4.
[note that the statement of MVT does hold here as

𝑓 (1)βˆ’π‘“ (βˆ’1)
1βˆ’(βˆ’1)

=

2βˆ’2
2

= 0 and 𝑓 β€² (0) = 0]

3. 𝑓 (π‘₯ ) = 2 ln (π‘₯ + 3) , π‘₯ ∈ [βˆ’2,0].
This elementary function is defined, continuous and differentiable on ( βˆ’3, ∞) so it is
continuous on [βˆ’2,0] and differentiable on (βˆ’2,0).
2

𝑓 (𝑏)βˆ’π‘“ (π‘Ž)

2 ln 3βˆ’2 ln 1

To find 𝑐, determine 𝑓 β€² (𝑐 ) = 𝑐+3 and
=
= ln 3. Solve for 𝑐:
π‘βˆ’π‘Ž
2
2
𝟐 βˆ’ πŸ‘ π₯𝐧 πŸ‘
= ln 3 , 2 = 𝑐 ln 3 + 3 ln 3 , 𝑐 =
β‰ˆ βˆ’1.18 ∈ [βˆ’2,0].
𝑐+3
π₯𝐧 πŸ‘

4. 𝑓 (π‘₯ ) =

1
√4βˆ’π‘₯2

. It is defined where 4 βˆ’ π‘₯ 2 > 0, i.e. π‘₯ ∈ (βˆ’2,2) .

It is increasing for π‘₯ > 0, 𝒙 ∈ (𝟎, 𝟐) because π‘₯ 2 increases, 4 βˆ’ π‘₯ 2 decreases, √4 βˆ’ π‘₯ 2
decreases and

1
√4βˆ’π‘₯2

increases.

In turn, it is decreasing for π‘₯ < 0, 𝒙 ∈ ( βˆ’πŸ, 𝟎) because π‘₯ 2 decreases, 4 βˆ’ π‘₯ 2 increases,
√4 βˆ’ π‘₯ 2 increases and

1
√4βˆ’π‘₯2

decreases.

[We can also prove this using derivative:

1 β€²
3
1
2 )βˆ’ 2 )
)
((
𝑓 π‘₯ = 4βˆ’π‘₯
= βˆ’ (4 βˆ’ π‘₯ 2 )βˆ’2 βˆ™ (βˆ’2π‘₯ ) =
2
β€²(

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

The graph:

π‘₯

3,

(4 βˆ’ π‘₯ 2 )2

5. 𝑓 (π‘₯ ) = (π‘₯ βˆ’ 2)(π‘₯ βˆ’ 3)2 .
Find the derivative: 𝑓 β€² (π‘₯) =
= (π‘₯ βˆ’ 2) β€² (π‘₯ βˆ’ 3)2 + (π‘₯ βˆ’ 2)((π‘₯ βˆ’ 3)2 )β€² =
= (π‘₯ βˆ’ 3) 2 + 2(π‘₯ βˆ’ 2)(π‘₯ βˆ’ 3) =
= (π‘₯ βˆ’ 3)(π‘₯ βˆ’ 3 + 2π‘₯ βˆ’ 4) =
= (π‘₯ βˆ’ 3)(3π‘₯ βˆ’ 7)...

Borys_S (7932)
Boston College

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