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Math-106, Fall 2004, MT-2 (Type A) Problems and Solutions
( ) (^4 44) ( 1 ) 4 ( 1 ) ( 1 )
3 2 f ′^ x = x − x = x ⋅ x − = x ⋅ x − ⋅ x +
For x ∈(− ∞,− 1 ), f ′( x ) < 0 ⇒ f is decreasing.
For x ∈( − 1 , 0 ), f ′( )^ x > 0 ⇒ f is increasing.
For x ∈( 0 , 1 ), f ′( ) x < 0 ⇒ f is decreasing.
For x ∈( 1 , ∞ ), f ′( )^ x > 0 ⇒ f is increasing.
f ′ ( ) x = 0 ⇒ x = 0 or x = 1 or x =− 1.
x =− 1 : f ′^ changes sign from – to + ⇒ It is local minimum with value f ( − 1 ) = 2.
x = 0 : f ′^ changes sign from + to – ⇒ It is local maximum with value f ( ) 0 = 3.
x = 1 : f ′^ changes sign from – to + ⇒ It is local minimum with value f ( ) 1 = 2.
( ) 12 4 4 ( 3 1 )
2 2 f ′′^ x = x − = ⋅ x −
For ′′( ) > ⇒
x , f x concave up.
For ′′( ) < ⇒
x f x concave down.
For ′′( ) > ⇒
x f x concave up.
( ) 3
f ′′ x = 0 ⇒ x =± and when
3
x = ± , f changes sign. In other words
and
are both inflection points.
If we put
x x u e e
− = + then du ( e e ) dx
x x = − ⋅
− and
∫
tanh x ⋅ dx = u C ( e e ) C ( x ) C u
du (^) x x = + = + + = + ′
− ∫
ln ln lncosh.
Solution
F ( ) x f ( ) t dt
x
a
∫
( ) ( ( ) ( )) ( ) ( ) (^) =
∫ ∫
→ →
x h x
h h
f t dt f t dt h
F x h F x h
F x
0 0
0 0
lim
lim
= ( ) ( ) ( ) (^)
∫ ∫ ∫
→
x xh x
x
h
f t dt f t dt f t dt (^0) h (^) 0 0
lim = ( ) ∫
→
x h
x
h
f t dt h
lim 0
By the Mean Value Theorem for Integrals, there exists c ∈ ( x , x + h )such that
( ) ∫
⋅ ⋅
x h
x
f t dt h
= f ( c ).
Since f is a continuous function, we have F ( x ) f ( ) c f ( ) c f ( ) x h c x
→ →
lim lim 0
( )
( )
( )
( x )
x x t x
dt
dt
d
x
2
2 2
3
1 sin
ln 1 sin
sin 2 1 sin ln 1 sin
2 ln +
∫
Solution : We have dx
dx
dy L (^) ⋅
∫
2
1
2
x
x
x x
x x
dx
dy
So 2
(^2222)
x
x x
x
x
dx
dy − + = +
x
x
x
x
x
x
x
x x
2 2 2 2
2
4 2
=
since x^ ∈[^1 ,^2 ], so x ≥ 0.
∴ (^) =
∫ ∫ ∫ 1
2 2 2
1
2
1
2
1
2
ln 2 2
x
x dx x
dx x x
x dx x
x L
= ln 2 4
ln 2 2
ln 1 2
ln 2 2
First solution:
[ ( )] ( )
−
− ^ + +
(^) + + 1 2 2
ln 1 ln 1
2 1 1 2
sinhln 1
2 2
x x x x
e e x x
x x x x
2
2 2 2
2
2 2
x x
x x x x
x x
x x
= x ( ( ) x )
x x
x x x 1
2
2 2
sinhsinh
1
[ ( x x )] [ ( ) x ]
2 1 sinh ln 1 sinhsinh
− ⇒ + + =
⇒ ( [ ( )]) [ ( ( ))]
↓
− −
↓
− x + x + = x
1 2 1 1 sinh sinhln 1 sinh sinhsinh
ln ( 1 )
2 x + x + = ( x )
1 sinh
− .
Second solution:
( )
( )
( ) ( )
2 sinh
2 sinh
sinh
1
1 sinh
cosh
sinh
sinh 1 1
1
y x y x
x
y dx
d
x dx
d
y y x
y
− −
−
= =
=
−
[ ]
ln 1 2 2
2
2
2
2 2
x x x
x
x x
x x
x
x
x x dx
d
