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Practice
Extremal value problems
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Extremal value problems: Problems with Solutions
Problem 1
Find the minimal value of the function [tex]f(x)=3x^3-9x^2+6[/tex] for [tex]x \in [-1;5][/tex].
Solution:
The minimal value could be in both ends of the interval or in a point, in which there is a local minimum for the function. We calculate [tex]f(-1)=3.(-1)-9+6=-6[/tex] and [tex]f(5)=3.5^3-9.5^2+6=5^2(3.5-9)+6=25.6+6>-6[/tex], which is obviously not the minimal value.
All that is left is to find the local extrema of the function. [tex]f'(x)=9x^2-18x=9x(x-2)[/tex], so there are extrema in [tex]x=0[/tex] and [tex]x=2[/tex]. [tex]f(0)=3.0-9.0+6=6[/tex], so it is not a minimal value. [tex]f(2)=3.8-9.4+6=-6[/tex]. So the minimal value is [tex]-6[/tex] and is reached in two points - [tex]x=-1[/tex] and [tex]x=2[/tex].
Problem 2
Find the maximum value of the function [tex]f(x)=x-5[/tex] if
x
is a number between
-5
and
13
.
Solution:
Since
f(x)
is a linear function whose slope is
1
, a positive number, it is strictly increasing for all
x
. Therefore its maximal value is reached for the largest value of
x
,
x=13
and
f(x)=13-5=8
.
Problem 3
Find the maximal value for the function [tex]f(x)=4sin(x)[/tex]
Solution:
It is known that [tex]-1 \le sin(x) \le 1[/tex]. We multiply this by 4 and get [tex]-4 \le 4sin(x) \le 4[/tex], so the maximal value for this function is 4.
Problem 4
Find the minimal value for the function [tex]f(x)=|x|-1[/tex].
Solution:
[tex]|x| \ge 0[/tex], we add
-1
to both sides of the inequality and get [tex]|x|-1 \ge -1[/tex].
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