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Practice: Absolute Extrema on Closed Interval
Related Topics
Wize University Calculus 1 Textbook > Applications of Differentiation
Maximum and Minimum on Closed Intervals
4 Activities
Wize University Calculus 1 Textbook > Applications of Differentiation
1st and 2nd Derivative Tests
4 Activities
Q.
\textbf{Q.}
Q.
Find the absolute maximum and minimum values of
f
(
x
)
=
6
x
4
3
−
3
x
1
3
\displaystyle f(x)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}}
f
(
x
)
=
6
x
3
4
−
3
x
3
1
on the interval
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
and then everywhere, using the second derivative test.
Absolut minimum value on [-1,1], x-coordinate =
Absolut minimum value on [-1,1], y-coordinate =
Absolut maximum value on [-1,1], x-coordinate =
Absolut maximum value on [-1,1], y-coordinate =
I don't know
Check Submission
More Maximum and Minimum on Closed Intervals Questions:
Maximum and Minimum on Closed Intervals
Find the absolute max of the function
f
(
x
)
=
cos
x
⋅
e
−
x
f\left(x\right)=\cos x\cdot e^{-x}
f
(
x
)
=
cos
x
⋅
e
−
x
on
[
0
,
π
]
\left[0,\pi\right]
[
0
,
π
]
Closed Intervals: Maximum and Minimum
Find the local and absolute extreme values of
f
(
x
)
=
3
x
−
4
x
2
+
1
f(x)=\frac{3x-4}{x^2+1}
f
(
x
)
=
x
2
+
1
3
x
−
4
on the interval
[
−
2
,
2
]
[-2,2]
[
−
2
,
2
]
Relative Extrema
Find and classify all local and absolute extrema of
f
(
x
)
=
sin
x
cos
x
f\left(x\right)=\sin x\cos x
f
(
x
)
=
sin
x
cos
x
on the interval
[
0
,
2
π
]
\left[0,\ \ 2\pi\right]
[
0
,
2
π
]
.
Concavity and Inflection Points: Graph of a Derivative
Practice: Graph of a Derivative
Suppose that
f
f
f
is a differentiable function on the interval
[
−
4
,
2
]
\left[-4,2\right]
[
−
4
,
2
]
. If the following graph represents the derivative of
f
f
f
, which of the following statements is correct?
Determine where the absolute extrema of
f
(
x
)
=
3
x
2
3
−
2
x
f(x)=3x^{\frac{2}{3}}-2x
f
(
x
)
=
3
x
3
2
−
2
x
on the interval
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
occur, and their values.
Practice: Absolute Extrema on Closed Interval
Q.
\textbf{Q.}
Q.
Find the absolute maximum and minimum values of
f
(
x
)
=
6
x
4
3
−
3
x
1
3
\displaystyle f(x)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}}
f
(
x
)
=
6
x
3
4
−
3
x
3
1
on the interval
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
and then everywhere, using the second derivative test.
Practice: Relative Extrema with Trig
Q.
\textbf{Q.}
Q.
Find all the relative extrema of
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
\displaystyle f(x)=2e^x(\sin{x}-\cos{x})
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
on the interval
[
π
2
,
π
]
\bigg[\dfrac{\pi}{2},\pi\bigg]
[
2
π
,
π
]
.
Q.
\textbf{Q.}
Q.
Find the absolute extrema of
f
(
x
)
=
2
x
x
2
+
1
\displaystyle f(x)=\frac{2x}{x^2+1}
f
(
x
)
=
x
2
+
1
2
x
on the interval
[
0
,
2
]
[0,2]
[
0
,
2
]
. Then find the local/absolute extrema on its domain, and intervals of increase and decrease.
Practice: Graph of a Derivative (~F2018 Final Q17) (~F2016 Final Q16)
Practice: Graph of a Derivative
Suppose that
f
f
f
is a differentiable function on the interval
(
−
4
,
2
)
(-4,2)
(
−
4
,
2
)
. If the following graph represents the derivative of
f
f
f
, which of the following statements is correct?
Maximum and Minimum on Closed Intervals
Find the absolute max of the function
f
(
x
)
=
cos
x
⋅
e
−
x
f\left(x\right)=\cos x\cdot e^{-x}
f
(
x
)
=
cos
x
⋅
e
−
x
on
[
0
,
π
]
\left[0,\pi\right]
[
0
,
π
]
Critical Points and Extrema: Maximum and Minimum on Closed Intervals
Find the absolute maximum of value of
f
(
x
)
=
2
x
x
2
+
1
\displaystyle f(x)=\frac{2x}{x^2+1}
f
(
x
)
=
x
2
+
1
2
x
on the interval
[
0
,
2
]
[0,2]
[
0
,
2
]
.
Critical points and extrema: Maximum and minimum on closed intervals
Find the absolute maximum value of
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
f(x)=2e^x(\sin x-\cos x)
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
on the interval
[
π
/
2
,
π
]
[\pi/2,\pi]
[
π
/2
,
π
]
.
Critical points and extrema: Maximum and Minimum on closed intervals
Find the absolute minimum of
f
(
x
)
=
arcsin
(
x
2
)
f(x)=\text{arcsin}\left(\frac{x}{2}\right)
f
(
x
)
=
arcsin
(
2
x
)
on the interval
[
1
,
2
]
[1,2]
[
1
,
2
]
.
Relative Extrema
Find and classify all local and absolute extrema of
f
(
x
)
=
sin
x
cos
x
f\left(x\right)=\sin x\cos x
f
(
x
)
=
sin
x
cos
x
on the interval
[
0
,
2
π
]
\left[0,\ \ 2\pi\right]
[
0
,
2
π
]
.
Find the absolute maximum value of
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
f(x)=2e^x(\sin x-\cos x)
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
on the interval
[
π
/
2
,
π
]
[\pi/2,\pi]
[
π
/2
,
π
]
.
Find the absolute minimum and absolute maximum of
f
(
x
)
=
x
1
3
(
7
−
x
)
2
f\left(x\right)=x^{\frac{1}{3}}\left(7-x\right)^2\
f
(
x
)
=
x
3
1
(
7
−
x
)
2
on
[
−
1
,
7
]
\left[-1,7\right]
[
−
1
,
7
]
.
Practice: Relative Extrema
Practice Question: Relative Extrema
Find and classify all local and absolute extrema of
f
(
x
)
=
sin
x
cos
x
f\left(x\right)=\sin x\cos x
f
(
x
)
=
sin
x
cos
x
on the interval
[
0
,
2
π
]
\left[0,\ \ 2\pi\right]
[
0
,
2
π
]
.
Closed Intervals: Maximum and Minimum
Find the local and absolute extreme values of
f
(
x
)
=
3
x
−
4
x
2
+
1
f(x)=\frac{3x-4}{x^2+1}
f
(
x
)
=
x
2
+
1
3
x
−
4
on the interval
[
−
2
,
2
]
[-2,2]
[
−
2
,
2
]
The absolute maximum value of
f
(
x
)
=
1
−
x
2
3
x
+
5
f(x)=\dfrac{1-x^2}{3x+5}
f
(
x
)
=
3
x
+
5
1
−
x
2
, over the closed interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
is:
final114
The absolute maximum value of
f
(
x
)
=
1
−
x
2
3
x
+
5
f(x)=\dfrac{1-x^2}{3x+5}
f
(
x
)
=
3
x
+
5
1
−
x
2
, over the closed interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
is:
final114
The absolute maximum value of
f
(
x
)
=
1
−
x
2
3
x
+
5
f(x)=\dfrac{1-x^2}{3x+5}
f
(
x
)
=
3
x
+
5
1
−
x
2
, over the closed interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
is:
Find the absolute minimum of
f
(
x
)
=
arcsin
(
x
2
)
f(x)=\text{arcsin}\left(\frac{x}{2}\right)
f
(
x
)
=
arcsin
(
2
x
)
on the interval
[
1
,
2
]
[1,2]
[
1
,
2
]
.
Practice: Relative Extrema with Trig
Q.
\textbf{Q.}
Q.
Find all the relative extrema of
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
\displaystyle f(x)=2e^x(\sin{x}-\cos{x})
f
(
x
)
=
2
e
x
(
sin
x
−
cos
x
)
on the interval
[
π
2
,
π
]
\bigg[\dfrac{\pi}{2},\pi\bigg]
[
2
π
,
π
]
.
Q.
\textbf{Q.}
Q.
Find the absolute extrema of
f
(
x
)
=
2
x
x
2
+
1
\displaystyle f(x)=\frac{2x}{x^2+1}
f
(
x
)
=
x
2
+
1
2
x
on the interval
[
0
,
2
]
[0,2]
[
0
,
2
]
. Then find the local/absolute extrema on its domain, and intervals of increase and decrease.
Determine where the absolute extrema of
f
(
x
)
=
3
x
2
3
−
2
x
f(x)=3x^{\frac{2}{3}}-2x
f
(
x
)
=
3
x
3
2
−
2
x
on the interval
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
occur, and their values.
The absolute maximum value of
f
(
x
)
=
1
−
x
2
3
x
+
5
f(x)=\dfrac{1-x^2}{3x+5}
f
(
x
)
=
3
x
+
5
1
−
x
2
, over the closed interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
is:
Find the absolute minimum and absolute maximum of
f
(
x
)
=
x
1
3
(
7
−
x
)
2
on
[
−
1
,
7
]
f\left(x\right)=x^{\frac{1}{3}}\left(7-x\right)^2\ \text{ on }\left[-1,7\right]
f
(
x
)
=
x
3
1
(
7
−
x
)
2
on
[
−
1
,
7
]
.
For the following function, determine all the local and global minimum/maximum and inflection points over the given interval
h
(
x
)
=
x
5
−
x
3
+
1
over
[
−
2
,
2
]
h(x)= x^5-x^3+1\quad \text{over}\ [-2,2]
h
(
x
)
=
x
5
−
x
3
+
1
over
[
−
2
,
2
]
For the following function, determine all the local and global minimum/maximum and inflection points over the given interval
f
(
x
)
=
x
3
−
3
x
2
over
[
−
1
,
3
]
f(x)=x^3-3x^2\quad \text{ over }\ [-1,3]
f
(
x
)
=
x
3
−
3
x
2
over
[
−
1
,
3
]
Find the absolute maximum of value of
f
(
x
)
=
2
x
x
2
+
1
\displaystyle f(x)=\frac{2x}{x^2+1}
f
(
x
)
=
x
2
+
1
2
x
on the interval
[
0
,
2
]
[0,2]
[
0
,
2
]
.
Find the absolute minimum and absolute maximum of
f
(
x
)
=
x
1
3
(
7
−
x
)
2
f\left(x\right)=x^{\frac{1}{3}}\left(7-x\right)^2\
f
(
x
)
=
x
3
1
(
7
−
x
)
2
on
[
−
1
,
7
]
\left[-1,7\right]
[
−
1
,
7
]
.
Concavity and Inflection Points: Graph of a Derivative
Practice: Graph of a Derivative
Suppose that
f
f
f
is a differentiable function on the interval
[
−
4
,
2
]
\left[-4,2\right]
[
−
4
,
2
]
. If the following graph represents the derivative of
f
f
f
, which of the following statements is correct?
More 1st and 2nd Derivative Tests Questions:
Critical points and extrema
Find all local extreme values of the function
f
(
x
)
=
x
3
−
6
x
2
+
9
x
+
1
f(x)=x^3-6x^2+9x+1
f
(
x
)
=
x
3
−
6
x
2
+
9
x
+
1
Derivative tests
Let
f
(
x
)
=
−
x
3
+
x
2
+
x
f(x) = -x^3+x^2+x
f
(
x
)
=
−
x
3
+
x
2
+
x
and use the first derivative test to find and classify the extreme values.
Since
f
is a negative cubic and we’re not given an interval,
f
has no absolute max or min. Here,
f
’
(
x
)
=
−
3
x
2
+
2
x
+
1
f’(x)=-3x^2+2x+1
f
’
(
x
)
=
−
3
x
2
+
2
x
+
1
, which exists everywhere. Let’s set it equal to zero and solve for
x
:
0
=
−
3
x
2
+
2
x
+
1
=
−
(
3
x
+
1
)
(
x
−
1
)
→
x
=
1
,
−
1
/
3.
0 = -3x^2+2x+1 = -(3x+1)(x-1) \rightarrow x=1,-1/3.
0
=
−
3
x
2
+
2
x
+
1
=
−
(
3
x
+
1
)
(
x
−
1
)
→
x
=
1
,
−
1/3.
Critical points and Extrema
Given this table of values, which of the following statements must NOT be true?
f
(
x
)
f
′
(
x
)
f
′
′
(
x
)
x
=
0
3
1
3
x
=
1
0
0
−
1
x
=
2
0
−
2
0
\begin{array}{|c|c|c|c|} \hline &f(x)&f'(x)&f''(x)\\ \hline x=0&3&1&3\\ x=1&0&0&-1\\ x=2&0&-2&0\\ \hline \end{array}
x
=
0
x
=
1
x
=
2
f
(
x
)
3
0
0
f
′
(
x
)
1
0
−
2
f
′′
(
x
)
3
−
1
0
Absolute Extrema on Closed Interval
Find the absolute maximum of
f
(
x
)
=
3
x
4
−
4
x
3
\displaystyle f(x)=3x^{4}-4x^{3}
f
(
x
)
=
3
x
4
−
4
x
3
on the interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
using the second derivative test.
Absolute Extrema on Closed Interval
Find the absolute extrema of
f
(
x
)
=
3
x
4
−
4
x
3
\displaystyle f(x)=3x^{4}-4x^{3}
f
(
x
)
=
3
x
4
−
4
x
3
on the interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
using the second derivative test.
Relative Extrema
Find and classify all local and absolute extrema of
f
(
x
)
=
sin
x
cos
x
f\left(x\right)=\sin x\cos x
f
(
x
)
=
sin
x
cos
x
on the interval
[
0
,
2
π
]
\left[0,\ \ 2\pi\right]
[
0
,
2
π
]
.
Critical Points and Extrema: Max and Min
Find and classify the critical points of
f
(
x
)
=
x
e
x
−
e
x
f(x)=xe^{x}-e^{x}
f
(
x
)
=
x
e
x
−
e
x
as relative maxima or minima.
Practice: Absolute Extrema on Closed Interval
Q.
\textbf{Q.}
Q.
Find the absolute maximum and minimum values of
f
(
x
)
=
6
x
4
3
−
3
x
1
3
\displaystyle f(x)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}}
f
(
x
)
=
6
x
3
4
−
3
x
3
1
on the interval
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
and then everywhere, using the second derivative test.
Absolute Extrema on Closed Interval
Find the absolute extrema of
f
(
x
)
=
3
x
4
−
4
x
3
\displaystyle f(x)=3x^{4}-4x^{3}
f
(
x
)
=
3
x
4
−
4
x
3
on the interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
using the second derivative test.
Continuity
Let
f
(
x
)
f(x)
f
(
x
)
be a continuous function on the open interval
(
a
,
b
)
(a, b)
(
a
,
b
)
. Which of the following four statements are always true.
Select all that apply.
Derivative Tests
Which of the following five functions are concave up on their whole domain? Select all that apply.
Critical points and Extrema
Given this table of values, which of the following statements must NOT be true?
f
(
x
)
f
′
(
x
)
f
′
′
(
x
)
x
=
0
3
1
3
x
=
1
0
0
−
1
x
=
2
0
−
2
0
\begin{array}{|c|c|c|c|} \hline &f(x)&f'(x)&f''(x)\\ \hline x=0&3&1&3\\ x=1&0&0&-1\\ x=2&0&-2&0\\ \hline \end{array}
x
=
0
x
=
1
x
=
2
f
(
x
)
3
0
0
f
′
(
x
)
1
0
−
2
f
′′
(
x
)
3
−
1
0
Relative Extrema
Find and classify all local and absolute extrema of
f
(
x
)
=
sin
x
cos
x
f\left(x\right)=\sin x\cos x
f
(
x
)
=
sin
x
cos
x
on the interval
[
0
,
2
π
]
\left[0,\ \ 2\pi\right]
[
0
,
2
π
]
.
Critical Points and Extrema
Let
f
(
x
)
f(x)
f
(
x
)
be a smooth function (a function that has infinitely many derivatives that exist everywhere) and consider
x
=
a
x = a
x
=
a
. Which of the following statements are correct? Check off all correct statements.
Practice
f
(
x
)
=
1
x
2
+
1
\displaystyle f(x)=\frac{1}{x^2+1}
f
(
x
)
=
x
2
+
1
1
.
The first and the second derivatives are
f
′
(
x
)
=
−
2
x
(
x
2
+
1
)
2
and
f
′
′
(
x
)
=
6
x
2
−
2
(
x
2
+
1
)
3
.
\boxed{f^{\prime}(x)=\frac{-2x}{\left(x^{2}+1\right)^{2}}\quad\text{ and }\quad f^{\prime\prime}(x)=\frac{6x^{2}-2}{\left(x^{2}+1\right)^3}.}
f
′
(
x
)
=
(
x
2
+
1
)
2
−
2
x
and
f
′′
(
x
)
=
(
x
2
+
1
)
3
6
x
2
−
2
.
Practice
f
(
x
)
=
1
x
2
+
1
\displaystyle f(x)=\frac{1}{x^2+1}
f
(
x
)
=
x
2
+
1
1
.
The first and the second derivatives are
f
′
(
x
)
=
−
2
x
(
x
2
+
1
)
2
and
f
′
′
(
x
)
=
6
x
2
−
2
(
x
2
+
1
)
3
.
\boxed{f^{\prime}(x)=\frac{-2x}{\left(x^{2}+1\right)^{2}}\quad\text{ and }\quad f^{\prime\prime}(x)=\frac{6x^{2}-2}{\left(x^{2}+1\right)^3}.}
f
′
(
x
)
=
(
x
2
+
1
)
2
−
2
x
and
f
′′
(
x
)
=
(
x
2
+
1
)
3
6
x
2
−
2
.
Practice: Minimum Value (Similar to April 2018 Q45)
Find the minimum value of
f
(
x
)
=
1
4
x
4
+
1
3
x
3
f(x)=\frac{1}{4}x^4+\frac{1}{3}x^3
f
(
x
)
=
4
1
x
4
+
3
1
x
3
Practice: Max, Min, and Inflection Points
Practice Question: Max, Min, and Inflection Points
Given the following table of values, which statement(s) must
always
be true about
f
(
x
)
f\left(x\right)
f
(
x
)
?
i.)
f
(
x
)
f\left(x\right)
f
(
x
)
has 3 critical points
Derivative tests
Let
f
(
x
)
=
−
x
3
+
x
2
+
x
f(x) = -x^3+x^2+x
f
(
x
)
=
−
x
3
+
x
2
+
x
and use the first derivative test to find and classify the extreme values.
Since
f
is a negative cubic and we’re not given an interval,
f
has no absolute max or min. Here,
f
’
(
x
)
=
−
3
x
2
+
2
x
+
1
f’(x)=-3x^2+2x+1
f
’
(
x
)
=
−
3
x
2
+
2
x
+
1
, which exists everywhere. Let’s set it equal to zero and solve for
x
:
0
=
−
3
x
2
+
2
x
+
1
=
−
(
3
x
+
1
)
(
x
−
1
)
→
x
=
1
,
−
1
/
3.
0 = -3x^2+2x+1 = -(3x+1)(x-1) \rightarrow x=1,-1/3.
0
=
−
3
x
2
+
2
x
+
1
=
−
(
3
x
+
1
)
(
x
−
1
)
→
x
=
1
,
−
1/3.
Critical points and extrema
Find all local extreme values of the function
f
(
x
)
=
x
3
−
6
x
2
+
9
x
+
1
f(x)=x^3-6x^2+9x+1
f
(
x
)
=
x
3
−
6
x
2
+
9
x
+
1
Absolute Extrema on Closed Interval
Find the absolute extrema of
f
(
x
)
=
3
x
4
−
4
x
3
\displaystyle f(x)=3x^{4}-4x^{3}
f
(
x
)
=
3
x
4
−
4
x
3
on the interval
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
using the second derivative test.
Critical Points and Extrema: Max and Min
Find and classify the critical points of
f
(
x
)
=
x
e
x
−
e
x
f(x)=xe^{x}-e^{x}
f
(
x
)
=
x
e
x
−
e
x
as relative maxima or minima.
Derivative Tests
Find the local extreme values of
h
(
x
)
=
x
4
−
2
x
2
h(x)=x^4-2x^2
h
(
x
)
=
x
4
−
2
x
2
using the second derivative test.