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MATH 115
QUIZ5-SAMPLE
November 7, 2016
Absolute min and max on a closed interval. Graph a function with min/max, inflection
points, asymptotes and x/y intercepts.
Section 4.3: 33-36 and Section 4.3:37-60. Section 4.4:9-38
5x
find the following information:
(x − 1)2
(i) Domain.
1. For f (x) =
(ii) x-intercept and y-intercept.
(iii) Asymptotes.
(iv) Intervals of increasing and decreasing.
(v) Relative extremum.
Solution:
(i) Domain can be found by excluding the zeros of the denominator.
(x − 1)2 = 0 gives x = 1 repeated.
Use number line to see that the domain is (−∞, 1) ∪ (1, ∞) .
(ii) x-intercepts are solutions to the equation y = 0. In this case
5x
= 0 which gives 5x = 0
(x − 1)2
and x=0 .
y-intercepts are result of plugging in x = 0 in the function: y =
5(0)
= 0.
(0 − 1)2
y=0 Notice that the two intercepts coincide.
(iii) Vertical asymptotes are the solutions to setting the denominator equal to zero.
(x − 1)2 = 0 gives x = 1 .
Horizontal asymptotes are non-infinite behaviours at infinities.
y = lim f (x) = 0 and also y = lim f (x) = 0
x→∞
x→−∞
So the horizontal asymptote is y=0 .
(iv) Find f ′ (x) =
5(x − 1)2 − 2(x − 1)(5x) 5x2 − 10x + 5 − 10x2 + 10x
=
=
((x − 1)2 )2
(x − 1)4
5(−x2 + 1) 5(−x + 1)(x + 1) −5(x + 1)
=
=
(x − 1)4
(x − 1)4
(x − 1)3
To find the points where the derivative does not exists set the denominator equal to zero and get
x = 1 which is not in the domain.
Setting f ′ = 0 to get −5(x + 1) = 0 which is x = −1 the only critical number.
f′
−−
0
−1
+
−−
1
Increasing on (−1, 1) and decreasing on (−∞, −1) and (1, ∞)
(v) The only critical number is x = −1 and according to increasing decreasing picture, a local
minimum is at x = −1. Find the value of the local minimum by plugging it in the original
5(−1)
= −5/4 .
(−1−1)2
5
find this information:
(x + 2)2
(i) Domain.
2. For f (x) =
(ii) x-intercept and y-intercept.
(iii) Asymptotes.
(iv) Intervals of increasing and decreasing.
(v) Relative extremum.
(vi) Intervals of concavity.
Solution: (i) Domain can be found by excluding the zeros of the denominator.
(x + 2)2 = 0 gives x = −2 repeated.
Use number line to see that the domain is (−∞, −2) ∪ (−2, ∞) .
(ii) x-intercepts are solutions to the equation y = 0. In this case
5
= 0 which is never possible
(x + 2)2
x-intercept DNE .
find y-intercept by plugging in x = 0.
y=
5
(0+2)2
= 5/4
(iii) Vertical asymptote: By setting the denominator equal to zero.
x + 2 = 0 gives x = −2 .
Horizontal asymptote by finding the limit as x → infinities.
lim f (x) = 0 and lim f (x) = 0
x→∞
x→−∞
y=0 is the horizontal asymptote.
(iv) f ′′ (x) =
−10
.
(x + 2)3
It does not have a critical number. Since the derivative is never zero and the numbers where
the derivative does not exists are not in the domain. Use the point of discontinuity to make the
number line.
f′
+
+
−2
The function is increasing on interval (−∞, −2) and decreasing on interval (−2, ∞).
30
(vi) The second derivative is f ′′ (x) =
.
(x + 2)4
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Again use the point of discontinuity only since no critical number exists.
f′
+
−−
−2
Concave upward on intervals (−∞, −2) and (2, ∞).
No inflection point.
3. Find Absolute maxima and absolute minima over the given closed interval.
(a) f (x) = −x3 + 3x2 + 5 over the interval [1, 5].
Solution:
• Note that the function is a polynomial and hence continuous everywhere. By the Extreme value theorem, both extremum exist on a closed interval.
• Take the derivative f ′ (x) = −3x2 + 6x. f ′ exist everywhere so set f ′ = 0 to find the
critical numbers.
−3x2 + 6x = 0 gives x = 0 and x = 2.
• List all critical numbers in the interval and the end points. {1, 2, 5}.
• Evaluate the original functions at those points.
f (1) = −(1)3 + 3(1)2 + 5 = 7
f (2) = −(2)3 + 3(2)2 + 5 = 9(largest)
f (5) = −(5)3 + 3(5)2 + 5 = −45(smallest)
• So The absolute maxima is achieved at x = 2 and is 9
And absolute minima is -45 and is achieved at x = 5.
(b) f (x) = x8/3 − 4x2/3 over the interval [−8, 27].
Solution:
• The function is not a polynomial so it takes more effort to know if it is continuous. All
terms of this function are third roots and third roots are defined everywhere. So the
function is defined everywhere. If you graph the function, you won’t see any gaps or
holes. Therefore, it is continuous everywhere.
8
2
8
1
• Find the derivative: f ′ (x) = x5/3 − 4( )x−1/3 = ( )(x5/3 − 1/3 )
3
3
3
x
Now notice that f ′ is not defined at x = 0 ( x = 0 is a critical number), so set f ′ = 0 to
find the rest of the critical numbers.
(8/3)(x5/3 −
1
)
x1/3
=0
which means x5/3 −
1
=0
x1/3
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Use common denominator:
(x5/3 )(x1/3 ) − 1
=0
x1/3
x2 − 1
That is, 1/3 = 0.
x
Solving zeros of a fraction means solving the zeros of the numerator which are not
zeros of the denominator.
That is solving x2 − 1 = 0 which gives two more critical numbers x = −1 and x = 1.
• List all critical numbers in the interval and the end points.
{−8, −1, 0, 1, 27}
• Evaluate the original function f at all points in the list.
f (−8) = (−8)8/3 − 4(x2/3 ) = 240
f (−1) = (−1)8/3 − 4((−1)2/3 ) = −3 (smallest)
f (0) = (0)8/3 − 4(0)2/3 = 0
f (1) = (1)8/3 − 4(1)2/3 = −3 (smallest)
f (27) = (27)8/3 − 4(27)2/3 = 6525 (largest)
• The absolute maximum is 6525 at x = 27.
The absolute minimum is -3 at x = −1 and at x = 1.
1
1
(c) g(t) = ( 2
) over the interval [2, 4].
3 t −1
Solution:
• Is the function continuous over the interval?
The discontinuities of the function match the values that are not in the domain. Those
are the values that make the denominator zero. t2 − 1 = 0 which gives t = −1 and t = 1.
Function is continuous everywhere except at t = ±1.
Neither of these two are in the interval [2, 4] so the function is continuous over the
interval.
−t
2
.
• Take the derivative g ′ (t) = ( ) 2
3 (t − 1)2
g ′ is not defined at t = ±1 but those are neither in the domain nor in the interval. Set
g ′ = 0 again by setting the numerator equal to zero. Gives t = 0 as only critical number.
• List all end points and critical numbers is the interval:
{2, 4}
• Evaluate the original g at all values in the list.
1
g(2) = (1/3)( 4−1
) = 1/9 ( biggest number.)
1
g(4) = (1/3)( 16−1
) = 1/(45) ( smallest number.)
• The absolute max is 1/9 at t = 2.
The absolute min is 1/45 at t = 4.
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4. Sketch a graph of a function f with the following properties:
(i) f ′ (−1) = 0, f ′ (1) = 0 and f ′ (3) = 0.
(ii) f ′′ (x) > 0 on intervals (−∞, 0) and (2, ∞).
(iii) f ′′ (x) < 0 on interval (0, 2).
(iv) f ′′ (0) = f ′′ (2) = 0.
(v) f (0) = 1 and f (2) = 3.
(vi) lim f (x) = ∞ and lim f (x) = ∞.
x→∞
x→∞
Solution:
f
(1, 4)
4
(2, 3)
3
(3, 3)
2
1
−2.
0
−1.
−1
(−1, −1)
−2
−3
−4
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(0, 1)
1.
2.
3.
Short Notes:
This will not appear on the quiz.
• Finding vertical asymptote of a rational function
P (x)
Suppose f (x) =
is a rational function where P (x) and Q(x) are polynomials.
Q(x)
Then, the line x = a is a vertical asymptote if Q(a) = 0 and P (a) ≠ 0. ( Zeros of the denominator.)
• Horizontal asymptote
The line y = b is a horizontal asymptote of a function f if either lim f (x) = b or lim f (x) = b
x→∞
x→−∞
• In general the behavior at infinities is the limit as x → ±∞ infinities.
It could be a number ( including 0) which gives the horizontal asymptote or it could be ±∞.
• The Extreme Value Theorem
If a function is continuous on a closed interval [a, b], then f has both an absolute maximum and absolute
minimum value on [a, b].
• Finding the absolute Extrema of f on a closed interval [a, b]
1. Make sure that the function is continuous on the interval.
2. Find the critical numbers of f that lie in (a, b).
3. Compute the values of f at each Critical number and find f (a) and f (b).
4. Absolute max is largest number computed in (c) and absolute min is the smallest one.
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