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Find the absolute maximum of value of f(x)=2x/(x^2+1) on the interval [0,2].
Related Topics
Wize University Calculus 1 Textbook > Applications of Differentiation
Maximum and Minimum on Closed Intervals
4 Activities
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
]
.
Answer
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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: 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.
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 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?