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Derivative Tests
Related Topics
Wize University Calculus 1 Textbook > Applications of Differentiation
1st and 2nd Derivative Tests
4 Activities
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.
0,1,2
0,-1,1
0,-1,-1
1,2,3
I don't know
Check Submission
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
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.
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.