identify any maximum or minimum values on the interval $-2 leq x leq 3$.\n15. $f(x)=5 x-2$\n16. $f(x)=-2…

Explanation:

Step1: Analyze the function type

These are all linear functions of the form (y = mx + b) (except constant functions). For a non - constant linear function (y=mx + b), if (m>0), the function is increasing; if (m < 0), the function is decreasing. For a constant function (y = c), the function has no increase or decrease.

Step2: Evaluate the function at the endpoints of the interval ([-2,3])

For (f(x)=5x - 2) ((m = 5>0), increasing function)

• When (x=-2), (f(-2)=5\times(-2)-2=-10 - 2=-12)
• When (x = 3), (f(3)=5\times3-2=15 - 2 = 13)

For (f(x)=-2x - 4) ((m=-2<0), decreasing function)

• When (x=-2), (f(-2)=-2\times(-2)-4=4 - 4=0)
• When (x = 3), (f(3)=-2\times3-4=-6 - 4=-10)

For (f(x)=-5) (constant function)

(f(x)=-5) for all (x\in[-2,3])

For (f(x)=\frac{3}{2}x + 1) ((m=\frac{3}{2}>0), increasing function)

• When (x=-2), (f(-2)=\frac{3}{2}\times(-2)+1=-3 + 1=-2)
• When (x = 3), (f(3)=\frac{3}{2}\times3+1=\frac{9}{2}+1=\frac{9 + 2}{2}=\frac{11}{2}=5.5)

For (f(x)=\frac{2}{3}x-6) ((m=\frac{2}{3}>0), increasing function)

• When (x=-2), (f(-2)=\frac{2}{3}\times(-2)-6=-\frac{4}{3}-6=-\frac{4 + 18}{3}=-\frac{22}{3}\approx - 7.33)
• When (x = 3), (f(3)=\frac{2}{3}\times3-6=2 - 6=-4)

For (f(x)=\frac{1}{2}) (constant function)

(f(x)=\frac{1}{2}) for all (x\in[-2,3])

For (f(x)=-x + 3) ((m=-1<0), decreasing function)

• When (x=-2), (f(-2)=-(-2)+3=2 + 3=5)
• When (x = 3), (f(3)=-3 + 3=0)

For (f(x)=20-3x) ((m=-3<0), decreasing function)

• When (x=-2), (f(-2)=20-3\times(-2)=20 + 6=26)
• When (x = 3), (f(3)=20-3\times3=20 - 9=11)

Answer:

• Problem 15: The minimum value is (-12) (at (x = - 2)) and the maximum value is (13) (at (x = 3)).
• Problem 16: The maximum value is (0) (at (x=-2)) and the minimum value is (-10) (at (x = 3)).
• Problem 17: The function has a constant value of (-5) (both maximum and minimum).
• Problem 18: The minimum value is (-2) (at (x=-2)) and the maximum value is (5.5) (at (x = 3)).
• Problem 19: The minimum value is (-\frac{22}{3}) (at (x=-2)) and the maximum value is (-4) (at (x = 3)).
• Problem 20: The function has a constant value of (\frac{1}{2}) (both maximum and minimum).
• Problem 21: The maximum value is (5) (at (x=-2)) and the minimum value is (0) (at (x = 3)).
• Problem 22: The maximum value is (26) (at (x=-2)) and the minimum value is (11) (at (x = 3)).