To evaluate the expression [tex]\(\frac{16 - x^2}{6x}\)[/tex] within the interval [tex]\((-5, -3)\)[/tex], we need to examine the behavior of this expression at the endpoints of the interval and possibly within the interval. Here's the step-by-step solution:
1. Evaluate at the Endpoints:
- Calculate the value of the expression at [tex]\(x = -5\)[/tex].
- Calculate the value of the expression at [tex]\(x = -3\)[/tex].
2. Step-by-step Calculation:
i. For [tex]\(x = -5\)[/tex]:
[tex]\[
\frac{16 - (-5)^2}{6 \cdot (-5)} = \frac{16 - 25}{-30} = \frac{-9}{-30} = \frac{9}{30} = \frac{3}{10}
\][/tex]
ii. For [tex]\(x = -3\)[/tex]:
[tex]\[
\frac{16 - (-3)^2}{6 \cdot (-3)} = \frac{16 - 9}{-18} = \frac{7}{-18} = -\frac{7}{18}
\][/tex]
3. Evaluate Within the Interval:
- The expression [tex]\(\frac{16 - x^2}{6x}\)[/tex] is a rational function. Since the function is continuous and differentiable on the open interval [tex]\((-5, -3)\)[/tex], we do not expect any jumps or undefined points within this interval.
4. Behavior of the Function:
- As [tex]\(x\)[/tex] moves from [tex]\(-5\)[/tex] to [tex]\(-3\)[/tex], we have seen that the value at [tex]\(x = -5\)[/tex] is [tex]\(\frac{3}{10}\)[/tex].
- At [tex]\(x = -3\)[/tex], the value is [tex]\(-\frac{7}{18}\)[/tex].
5. Conclusion:
- Although we've calculated the values at the endpoints, the overall behavior of the expression within the interval should be evaluated more broadly by considering a few test points.
Let's pick an intermediate point [tex]\(x = -4\)[/tex]:
iii. For [tex]\(x = -4\)[/tex]:
[tex]\[
\frac{16 - (-4)^2}{6 \cdot (-4)} = \frac{16 - 16}{-24} = \frac{0}{-24} = 0
\][/tex]
These calculations summarize the behavior of the expression over the interval:
- At [tex]\(x = -5\)[/tex], the expression evaluates to [tex]\(\frac{3}{10}\)[/tex].
- At [tex]\(x = -4\)[/tex], the expression evaluates to [tex]\(0\)[/tex].
- At [tex]\(x = -3\)[/tex], the expression evaluates to [tex]\(-\frac{7}{18}\)[/tex].
Thus, the expression [tex]\(\frac{16 - x^2}{6x}\)[/tex] evaluates to [tex]\(\frac{3}{10}\)[/tex] at [tex]\(x = -5\)[/tex] and [tex]\(-\frac{7}{18}\)[/tex] at [tex]\(x = -3\)[/tex], transitioning through [tex]\(0\)[/tex] at [tex]\(x = -4\)[/tex] within the interval [tex]\((-5, -3)\)[/tex].
