Answers and Explanations 311
Answers
401–500
Because the function is concave down at x = π
3
, it has a local maximum there, and
because the function is concave up at x = 5
3
π , it has a local minimum there. To find the
corresponding y value on f
(x), substitute the critical numbers into the original
function:
f π
π
π
π
π
3
2
3
3
2
3
2
3
3 3
( ) ( )
=
−
=
−
=
−
sin
f 5
3
2
5
3
5
3
2
3
2
5
3
3 5
3
π
π
π
π
π
( ) ( )
=
−
= −
−
= −
−
sin
Therefore, the local maximum is at π
π
3
3 3
,
−
(
) , and the local minimum is at
5
3
3 5
3
π
π
, −
−
(
) .
481.
3
Recall Rolle’s theorem: If f is a function that satisfies the following three hypotheses:
✓ f is continuous on the closed interval [a, b]
✓ f is differentiable on the open interval (a, b)
✓ f (a) = f
(b)
then there is a number c in (a, b) such that f
'(c) = 0.
Notice that the given function is differentiable everywhere (and therefore continuous
everywhere) because it’s a polynomial. Then verify that f (0) = f
(6) so that Rolle’s theorem
can be applied:
f
f
0 0 0 1 1
6 6 6 6 1 1
2
2
( ) = − + =
( ) = − ( )+ =
Next, find the derivative of the function, set it equal to zero, and solve for c:
′
f c
c
c
c
( ) = −
− =
=
2 6
2 6 0
3
Answers
401–500
Because the function is concave down at x = π
3
, it has a local maximum there, and
because the function is concave up at x = 5
3
π , it has a local minimum there. To find the
corresponding y value on f
(x), substitute the critical numbers into the original
function:
f π
π
π
π
π
3
2
3
3
2
3
2
3
3 3
( ) ( )
=
−
=
−
=
−
sin
f 5
3
2
5
3
5
3
2
3
2
5
3
3 5
3
π
π
π
π
π
( ) ( )
=
−
= −
−
= −
−
sin
Therefore, the local maximum is at π
π
3
3 3
,
−
(
) , and the local minimum is at
5
3
3 5
3
π
π
, −
−
(
) .
481.
3
Recall Rolle’s theorem: If f is a function that satisfies the following three hypotheses:
✓ f is continuous on the closed interval [a, b]
✓ f is differentiable on the open interval (a, b)
✓ f (a) = f
(b)
then there is a number c in (a, b) such that f
'(c) = 0.
Notice that the given function is differentiable everywhere (and therefore continuous
everywhere) because it’s a polynomial. Then verify that f (0) = f
(6) so that Rolle’s theorem
can be applied:
f
f
0 0 0 1 1
6 6 6 6 1 1
2
2
( ) = − + =
( ) = − ( )+ =
Next, find the derivative of the function, set it equal to zero, and solve for c:
′
f c
c
c
c
( ) = −
− =
=
2 6
2 6 0
3
