To solve this problem, we need to follow these steps:
1. Define the function [tex]\( f(x) \)[/tex]:
The function we're dealing with is [tex]\( f(x) = -\ln(6x^2 + 3) \)[/tex].
2. Compute the first derivative [tex]\( f'(x) \)[/tex]:
The derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex] can be found using the chain rule.
Let [tex]\( g(x) = 6x^2 + 3 \)[/tex]. Then [tex]\( f(x) = -\ln(g(x)) \)[/tex], so:
[tex]\[
f'(x) = \frac{d}{dx}[-\ln(g(x))] = -\frac{1}{g(x)} \cdot \frac{dg(x)}{dx}
\][/tex]
Now, compute the derivative of [tex]\( g(x) = 6x^2 + 3 \)[/tex]:
[tex]\[
\frac{dg(x)}{dx} = 12x
\][/tex]
Substituting back into the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[
f'(x) = -\frac{1}{6x^2 + 3} \cdot 12x = -\frac{12x}{6x^2 + 3}
\][/tex]
3. Evaluate the slope at [tex]\( x = 8 \)[/tex]:
We need to substitute [tex]\( x = 8 \)[/tex] into the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[
f'(8) = -\frac{12(8)}{6(8)^2 + 3}
\][/tex]
Calculate the denominator:
[tex]\[
6(8)^2 + 3 = 6 \cdot 64 + 3 = 384 + 3 = 387
\][/tex]
Now, substitute back into the fraction:
[tex]\[
f'(8) = -\frac{12 \cdot 8}{387} = -\frac{96}{387}
\][/tex]
Simplify the fraction:
[tex]\[
f'(8) \approx -0.248062015503876
\][/tex]
4. Round the result to the nearest hundredth:
The slope at [tex]\( x = 8 \)[/tex] rounded to the nearest hundredth is:
[tex]\[
f'(8) \approx -0.25
\][/tex]
Thus, the slope of [tex]\( f(x) = -\ln(6x^2 + 3) \)[/tex] at the point [tex]\( (8, -5.96) \)[/tex] is approximately [tex]\(-0.25\)[/tex].
