Question

identify any maximum or minimum values on the interval $-2 leq x leq 3$.

1. $f(x)=5 x-2$
2. $f(x)=-2 x-4$
3. $f(x)=-5$
4. $f(x)=\frac{3}{2} x+1$
5. $f(x)=\frac{2}{3} x-6$
6. $f(x)=\frac{1}{2}$
7. $f(x)=-x+3$
8. $f(x)=20-3 x$

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 \(x = 3\)).