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Review For Major Quiz 4 SOLUTIONS
1.) (a) Find the relative extrema of f by applying the first-derivative text; (b) determine the values of x at which the relative extrema occur; (c) determine the intervals on which f is increasing; (d) determine the intervals on which f is decreasing. f ( ) x = x^3 − 9 x^2 + 15 x − 5 f '( ) x = 3 x^2 − 18 x + 15 = 3( x − 1)( x −5) x = 1,5 are critical numbers.
Consider the number line in helping to determine when the derivative is positive or negative.
f '( ) x > 0 on the intervals ( −∞,1) and (5, ∞ ), so f ( ) x is increasing on those intervals f '( ) x < 0 on the interval (1,5) so f ( ) x is decreasing on this intervals f (1)is a relative maximum. f (5)is a relative minimum.
2.) Find a and b such that the function defined by f ( ) x = x^3^ + ax^2 + b will have a relative
extremum at (-1,3). ( 1)− 3 + a ( 1) − 2 + b = 3 a + b = 4
'( ) 3 2 2 3( 1) 2 2 ( 1) 0 2 3 0 3 2
f x = x + ax → − + a − = − a + = a =
3 5 4 2 2
− + b = b =
3.) Let f be a function that is continuous on a closed interval and differentiable on the open interval, if the derivative is negative for all values in the open interval, what can be said about the function? f ( ) x is decreasing on the open interval
4.) Find the absolute extrema of the function on the indicated interval, if there are any, and determine the values of x at which the absolute extrema occur.
(a) f ( ) x = x^4^ − 12 x^2 + 36 ; [-2,6]
f '( ) x = 4 x^3^ − 24 x = 4 ( x x 2 −6)
x = 0, ± 6 are critical numbers. x = − 6 is not in [-2,6]. f ( 2)− = 4 f (0) = 36 f ( 6) = 0 is an absolute minimum.
f ( 6 )= 900 is an absolute maximum.
x – 1
x – 5
– –^ +
(b)
3
x x
x x x f x ; [0,4)
Here is a graph of f ( x ).
1 2 3 4 5 6
2
4
6
8
10
x
y
We see ( )
2
lim 6 x −^ f^ x →
= , but f ( 2 ) = 2. So, f ( x )never attains the value of 6, and hence, f ( x )
does not have an absolute maximum on [0,4). Likewise, ( )
4
lim 10 x −^ f^ x →
= − , and since f ( x )never
attains the value of -10, it has no absolute minimum on [0,4).
5.) Find the absolute maximum and minimum values of f ( ) x = x^3^ + 2 x^2 + x − 1 on [−1, 1].
f '( ) x = 3 x^2 + 4 x + 1 = ( 3 x + 1 ) ( x + 1 )
x = − − are critical numbers.
f ( − 1 ) = − 1
f
^ − = −^ ≈ − is an absolute minimum
f ( 1 ) = 3 is an absolute maximum.
6.) State and prove the Mean-Value Theorem. If you are not familiar with it, check your book and notes.
7.) What does the Extreme-Value Theorem state? If you are not familiar with it, check your book and notes.
8.) Determine if f(x) satisfies the conditions of Rolle’s theorem on the indicated interval. If so, find a suitable value for c that satisfies the conclusion of Rolle’s theorem. Otherwise, state why the condition fails.
In this case, the conclusion of the Mean Value Theorem guarantees the existence of a number c
in the open interval (1,3). So, the value of c guaranteed by the Mean Value Theorem is
c =.
10.) Find constants a, b, and c such that the function f ( ) x = ax^3 + bx^2 + cx + d will have a
relative maximum at (−1, 1) and a relative minimum at (1, −1). f '( ) x = 3 ax^2 + 2 bx + c
f ( − 1 ) = 1 − a + b − c + d = 1
f ( 1 ) = − 1 a + b + c + d = − 1
f ' ( − 1 ) = 0 3 a − 2 b + c = 0
f ' 1 ( ) = 0 3 a + 2 b + c = 0
Adding the first two equations together gives 2 b + 2 d = 0, or d = − b ; adding the second two
together gives 6 a + 2 c = 0, or c = − 3 a. Substituting these relations into the first equation yields
1 3 1, or 2
− a + b + a − b = a = , giving
c = −. Substituting the these relations into the second
equation yields
3 2 0, or 0, giving 0. 2 2
b b d
−^ −^ =^ =^ =
11.) Use the first derivative test to determine the relative extrema of f ( ) x = 2 x^3^ − 3 x^2 − 72 x.
f '( ) x = 6 x^2 − 6 x = 6 ( x x −1)
f '( ) x > 0 on the intervals ( −∞, 0) and (1, ∞) , so f ( ) x is increasing on those intervals f '( ) x < 0 on the interval (0,1) so f ( ) x is decreasing on this intervals f (0)is a relative maximum. f (1)is a relative minimum.
12.) (a) Find the relative extrema of f; (b) determine the values of x at which the relative extrema occur; (c) determine the intervals on which f is increasing; (d) determine the intervals on which f is decreasing.
(i) 3
2
−
x
x x f x 2 2 2 2 2
x x x x x x x x f x x x x
x = 1, 5 are critical numbers.
0 1 x – 1
- + +^ 6x - (^) – (^) + + – +
+^1 -^^3 -^5 +
f '( ) x > 0 on the intervals ( −∞,1) and (5, ∞ ), so f ( ) x is increasing on those intervals f '( ) x < 0 on the intervals (1,3) and (3,5) so f ( ) x is decreasing on this intervals f (1)is a relative maximum. f (5)is a relative minimum.
(ii) 3
2
x
x f x 2 2 2 2 2
x x x x x f x x x
x = 0 is a critical number.
f '( ) x > 0 on the interval (0, ∞) , so f ( ) x is increasing on the interval. f '( ) x < 0 on the intervals ( −∞, 0) , so f ( ) x is decreasing on this interval. f (0)is a relative minimum.
(iii) f ( x )= 2 x^3 − 9 x^2 + 27
f '( ) x = 6 x^2 − 18 x = 6 ( x x −3)
f '( ) x > 0 on the intervals ( −∞, 0) and (3, ∞) , so f ( ) x is increasing on those intervals f '( ) x < 0 on the interval (0, 3) so f ( ) x is decreasing on this intervals f (0)is a relative maximum. f (3) is a relative minimum.
13.) Sketch the graph of a function with the following properties: f '( ) x > 0 when x < − 1 f '( ) x > 0 when x > 3 f '( ) x < 0 when −1 < x < 3 f ''( ) x < 0 when x < 2 f ''( ) x > 0 when x > 2
You are not ready for this problem yet. Disregard it.
13.) Find the absolute extrema of the function on the indicated interval, if there are any, and determine the values of x at which the absolute extrema occur. Draw a sketch of the graph of the function on the interval.
(a) f ( ) x = 5 x − 4 ; (-2, 5]
f '( ) x = 5
-^0 – (^3) + x^ –^3 + – + - (^) + (^) + 6x
-^0 +
No absolute maximum or minimum. (The function is not continuous. What happens at x =3?)
16.) Find the absolute extrema of the function on the indicated interval, if there are any, and determine the values of x at which the absolute extrema occur. Draw a sketch of the graph of the function on the interval.
if x
if x f x x ; [3,5]
2
x f x x x
^ =
f (5)is an absolute maximum.
17.) Find the absolute extrema of the function on the indicated interval, if there are any, and determine the values of x at which the absolute extrema occur. f ( ) x = 4 − 3 x ; (-1, 2] f '( ) x = − 3 f (2) = − 2 is an absolute minimum.
18.) Find the absolute extrema of the function on the indicated interval, if there are any, and determine the values of x at which the absolute extrema occur. Draw a sketch of the graph of the function on the interval.
x if x
x if x f x ; [-2,2]
x f x x x
^ −^ ≤^ <
≤^ ≤
f − '(1) = 2 f + '(1) = (2)(1) = 2 Since f (^) + '(1) = f (^) −'(1), f ' exists at 1. f ( 2)− = − 1 , f (2) = 8 , f (2)is an absolute maximum. f ( 2)− is an absolute minimum.
19.) Determine each of the following limits.
(a) 0
lim 1
x
x → x
0 0
lim ln 1 lim ln 1
x
x x x → (^) x → x
+^ =^ +
1
0 1
ln 1 5 lim x
x x
−
→ −
2 1 0 2
lim^1 x
x x x
− − → −
0 1
lim x → 1 5 x −
0
lim x 5
x → x
So, 0 0
lim 1 1
x
x e → x
+^ =^ =.
(b) 0
sin
lim
x tan
x
→ x
0 0
sin sin lim lim tan sin cos
x x
x x x^ x x
→ →
0
cos lim sin x sin
x x → x
0 lim cos x x →
(c) 1 2
lim
x 4 3
x
→ x x
1 2 1
lim lim x (^) 4 3 x 1 3
x x → (^) x x → x x
1
lim x → x 3
(d)
2
0 4
2cos 2
lim^3
x
x x
→ x
2
0 4 0 3
2 cos 2 2sin lim 3 lim 3 x x 4
x x x x → (^) x → x
0 2
2 cos lim 3 x 12
x → x
f (2) = 0 is an absolute minimum.
f (1) = 1
f (5) = 39 is an absolute maximum.
