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

University of British Columbia

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1,2: Mean Value Theorem requires π(π₯ ) be continuous on [π, π] and differentiable on (π, π).
1. π (π₯ ) = βπ₯ 2 + 1, π₯ β [β1,1].
This function is continuous on [β1,1] but not differentiable on entire (β1,1) .
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 β1 as β β 0 .

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

π (1)βπ (β1)
1β(β1)

=

2β2
2

= 0 but never

π β² (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)βπ (β1)
1β(β1)

=

2β2
2

= 0 and π β² (0) = 0]

3. π (π₯ ) = 2 ln (π₯ + 3) , π₯ β [β2,0].
This elementary function is defined, continuous and differentiable on ( β3, β) so it is
continuous on [β2,0] and differentiable on (β2,0).
2

π (π)βπ (π)

2 ln 3β2 ln 1

To find π, determine π β² (π ) = π+3 and
=
= ln 3. Solve for π:
πβπ
2
2
π β π π₯π§ π
= ln 3 , 2 = π ln 3 + 3 ln 3 , π =
β β1.18 β [β2,0].
π+3
π₯π§ π

4. π (π₯ ) =

1
β4βπ₯2

. It is defined where 4 β π₯ 2 > 0, i.e. π₯ β (β2,2) .

It is increasing for π₯ > 0, π β (π, π) because π₯ 2 increases, 4 β π₯ 2 decreases, β4 β π₯ 2
decreases and

1
β4βπ₯2

increases.

In turn, it is decreasing for π₯ < 0, π β ( βπ, π) because π₯ 2 decreases, 4 β π₯ 2 increases,
β4 β π₯ 2 increases and

1
β4βπ₯2

decreases.

[We can also prove this using derivative:

1 β²
3
1
2 )β 2 )
)
((
π π₯ = 4βπ₯
= β (4 β π₯ 2 )β2 β (β2π₯ ) =
2
β²(

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 β² (π₯) =
= (π₯ β 2) β² (π₯ β 3)2 + (π₯ β 2)((π₯ β 3)2 )β² =
= (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)...

Borys_S (7932)
Boston College
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