Extremal value problems: Problems with Solutions

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 \times (-1)-9+6=-6[/tex] and [tex]f(5)=3 \times 5^3-9 \times 5^2+6=5^2(3 \times 5-9)+6=25 \times 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 \times 0-9 \times 0+6=6[/tex], so it is not a minimal value. [tex]f(2)=3 \times 8-9 \times 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].

Solution:
Since [tex]f(x)[/tex] is a linear function whose slope is [tex]1[/tex], a positive number, it is strictly increasing for all [tex]x[/tex]. Therefore its maximal value is reached for the largest value of [tex]x[/tex], [tex]x=13[/tex] and [tex]f(x)=13-5=8[/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.

Solution:
[tex]|x| \ge 0[/tex], we add [tex]-1[/tex] to both sides of the inequality and get [tex]|x|-1 \ge -1[/tex].