Find the absolute extreme values of the function on the interval.
-$$h ( x ) = \frac { 1 } { 2 } x + 3 , - 3 \leq x \leq 4$$
A) absolute maximum is $- 1$ at $x = - 3$; absolute minimum is $\frac { 3 } { 2 }$ at $x = 4$
B) absolute maximum is $- 1$ at $x = - 4$; absolute minimum is $- 4$ at $x = 3$
C) absolute maximum is $- 1$ at $x = 4$; absolute minimum is $\frac { 3 } { 2 }$ at $x = - 3$
D) absolute maximum is 5 at $x = 4$; absolute minimum is $\frac { 3 } { 2 }$ at $x = - 3$

Question

Quizplus · Accepted Answer

The function $h(x) = \frac{1}{2}x + 3$ is a linear function, which means it increases or decreases steadily across its domain. To find the absolute extreme values on the interval $[-3, 4]$, we evaluate the function at the endpoints of the interval. At $x = -3$, $h(-3) = \frac{1}{2}(-3) + 3 = -\frac{3}{2} + 3 = \frac{3}{2}$. At $x = 4$, $h(4) = \frac{1}{2}(4) + 3 = 2 + 3 = 5$. Since the function is increasing over the interval, the absolute minimum is $\frac{3}{2}$ at $x = -3$, and the absolute maximum is $5$ at $x = 4$.

Find the Absolute Extreme Values of the Function on the Interval