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151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
November 12
1. Here is the graph of f 0 (x). The domain is given and it is all real numbers. Where is f (x)
increasing? decreasing? Classify the critical values of f (x) as relative min, relative max, or
neither.
The graph of f 0 (x)
E
A
B
C
D
This is the graph of f 0 (x). Our conclusions below are based on the fact that we are looking at
a picture of f 0 (x).
Since the function, f (x), is increasing when f 0 (x) is positive, then we look at the graph to see
where f 0 (x) is above the x-axis. (A, C), (C, D), (D, E).
Since the function, f (x), is decreasing when f 0 (x) is negative, then we look at the graph to see
where f 0 (x) is below the x-axis. (−∞, A), (E, ∞).
Now classify the critical values:
The critical values are A, C, D, and E.
To the left of A, the function is decreasing and to the right of A the function is increasing.
we have a relative min at x = A.
To the left of C, the function is increasing and to the right of C the function is increasing.
we have neither of these at x = C.
To the left of D, the function is increasing and to the right of D the function is increasing.
we have neither of these at x = D.
To the left of E, the function is increasing and to the right of E the function is decreasing.
we have a relative max at x = E.
Thus
Thus
Thus
Thus
2. Now lets redo number 1; but this time let it be the graph of f (x). The domain will be all real
numbers except x = D.
Where is f (x) increasing? decreasing? Classify the critical values of f (x) as relative min, relative
max, or neither.
This is the graph of f (x). Our conclusions below are based on the fact that we are looking at a
picture of f (x).
The function, f (x), is increasing on the intervals: (−∞, A), (C, D)
The function, f (x), is decreasing on the intervals: (B, C), (E, ∞).
Now classify the critical values:
The critical values are B and C.
There is a relative max at x = B.
There is a relative min at x = C.
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151 WebCalc Fall 2002-copyright Joe Kahlig
The graph of f (x)
E
B
A
C
D
3. Find where the function is increasing and decreasing and classify the critical values as rel max,
rel min, or neither.
√
(a) y = 3 x2 − x
First: the domain of the function is all real numbers. Now take a derivative and simplify
it.
2x − 1
y0 =
2
3(x2 − x) 3
Setting the derivative equal to zero and solving for x gives 2x − 1 = 0 or x = 12 . The other
critical values are found by looking to see where the derivative is undefined. The derivative
is undefined when x2 − x = 0 or x(x − 1) = 0. Thus x = 0 and x = 1. Since these values
are in the domain, they are also critical values.
−
−
+
+
y0
−1
inc:
0
1
2, 1
.1
, (1, ∞)
dec: (−∞, 0), 0, 12
0.5
.7
1
3
There is a relative min at x = 0. The other critical values, x = 0 and x = 1, are neither.
(b) y = x ln x
First: the domain is all real numbers greater than 0. Now take a derivative and simply it.
y 0 = ln(x) + 1
Setting the derivative equal to zero and solving for x gives ln x = −1 or x = e−1 ≈ .3678.
This is our only critical value. The derivative will be undefined for all values of x ≤ 0, but
none of these values are in the domain, so they are not critical values.
y0
not in
domain
−
0 .1
+
.367
2
inc: (e−1 , ∞)
dec: (0, e−1 )
There is a relative min at x = e−1 .
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151 WebCalc Fall 2002-copyright Joe Kahlig
4. If y 0 and the domain are given below, find where the function is increasing and decreasing and
classify the critical values as rel max, rel min, or neither.
2
(a) y 0 = x(x2 − 25)(x − 10)4 ex −9 and the domain is all real numbers.
2
Setting the derivative equal to zero and solving for x gives x = 0, −5, 5, 10. Note: ex −9
is always positive. There are no values of x where the derivative is undefined so we have
found all of the critical values.
y0
−
+
−6
−5
−1
−
1
0
+
5
7
+
10
15
inc (−5, 0), (5, 10), (10, ∞)
dec (−∞, −5), (0, 5)
There is a rel min at x = −5 and x = 5
There is a rel max at x = 0
There is neither of these at x = 10
−3x + 7
and the domain is all real numbers except x = −5.
(x + 5)3
Setting the derivative equal to zero and solving for x gives x = 73 . Note x = −5 is not a
critical value since it is not in the domain.
(b) y 0 =
y0
−
−7
−5
inc −5, 73
−
+
dec (−∞, −5),
7/3
0
7
3, ∞
5
There is a rel max at x =
7
3
There looks to be a rel min at x = −5. HOWEVER¡ THAT VALUE IS NOT A
CRITICAL VALUE SO IT CAN NOT BE A REL MIN. A function can change
direction and concavity at places that are not in the domain. That is why this value was
included on the sign chart.
5. Find the absolute maximum and absolute minimum for the function on the given intervals.
(a) y = f (x) = x3 − 12x on [−3, 5] This function is continuous and the interval given is a
closed interval. By the Extreme value theorem, we know that the abs max and abs min
will happen at critical values or at the ends of the interval.
The critical values for this function are x = −2 and x = 2. Now evaluate the function at
the critical values and at the ends of the interval.
f (−2) = 16, f (2) = −16, f (−3) = 9, and f (5) = 65
The absolute max is 65. The absolute min is -16.
Be careful about the question asked. The absolute max is 65 but it happens at x = 5.
(b) y = f (x) = x3 − 12x on [−3, 1] Note: same problem as above, but I changed the interval.
This function is continuous and the interval given is a closed interval. By the Extreme
value theorem, we know that the abs max and abs min will happen at critical values or at
the ends of the interval.
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151 WebCalc Fall 2002-copyright Joe Kahlig
The critical values for this function are x = −2 and x = 2. Note that the critical value of 2
is not in the interval, so ignore this value. Now evaluate the function at the critical values
and at the ends of the interval.
f (−2) = 16, f (1) = −11, and f (−3) = 9
The absolute max is 16. The absolute min is -11.
(c) y = x12 on [−2, 4]
This function is not continuous on the interval. so we must know what the graph does.
y 0 = x−2
− 3 and from the first derivative we can see that the function y is increasing on
the interval (−∞) and decreasing on the interval (0, ∞). [There are no critical values.]
Visualizing this and knowing that there is a vertical asymptote at x = 0, tell us that there
is not going to be an absolute max on this interval. The question is now what is the absolute
1
1
min? f (−2) = 14 and f (4) = 16
. so the absolute min is 16
.
(d) y = x12 on [2, 4] we know form part c that there are no critical values and that this function
is only discontinuous at x = 0, so we only need to look at the ends of the interval.
1
The absolute max is 14 and the abs min is 16
3π
(e) y = cos(x) on −π
Since the interval is not closed, we need to look at the graph. the
2 , 2
absolute max is 1 and the absolute min is -1.
6. For the function, f (x) = x1 , there is no number,c, in the interval [−2, 2] such that f 0 (c) = .25.
Why doesn’t this violate the mean value theorem?
Because the function is not continuous on the given interval.
7. A toll road is 100miles long. You clock into the start of the road at 8am and at 9:15am you exit
the end of the road. If the maximum speed allowed on the road is 70mph, do you get a ticket?
Justify your answer.
yes you get a ticket. By the mean value theorem, there is some spot on the roar where you had
100miles
an velocity of 1.25hour
= 80mph.
8. Suppose the f 0 (x) ≤ 2 on the interval [−3, 10] and f (1) = 1.5. What is the maximum value of
f (6)?
This is just an application of the tangent line. The tangent line At x = 1 is
y − f (1) = f 0 (1)(x − 1)
If we choose the largest value that f 0 (x) can be on this interval, which is 2, and put that in for
f 0 (1) we get this formula: y − 1.5 = 2(x − 1) Now just plug in a 6 for x and solve for y.
answer: the max value of f (6) is 11.5.