To find the critical points of the function $f(x) = \frac{9 - 13x}{13^x}$, we first need to compute its derivative $f'(x)$ and then set it equal to zero to find any critical points.

Step 1: Compute the derivative $f'(x)$

Using the quotient rule, which states that for a function $g(x) = \frac{u(x)}{v(x)}$, its derivative is given by: $g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$ we'll apply this to $f(x) = \frac{9 - 13x}{13^x}$.

Here, $u(x) = 9 - 13x$ and $v(x) = 13^x$. We need to find the derivatives $u'(x)$ and $v'(x)$:

• $u'(x) = -13$
• $v(x) = 13^x$ and $v'(x) = 13^x \ln(13)$ (since the derivative of $a^x$ is $a^x \ln(a)$).

Now apply the quotient rule: $f'(x) = \frac{(-13)(13^x) - (9 - 13x)(13^x \ln(13))}{(13^x)^2}$ Simplifying: $f'(x) = \frac{-13 \cdot 13^x - (9 - 13x) \cdot 13^x \ln(13)}{(13^x)^2}$ Factoring out $13^x$ in the numerator: $f'(x) = \frac{13^x \left[-13 - (9 - 13x)\ln(13)\right]}{(13^x)^2}$ Cancel $13^x$ in the numerator and denominator: $f'(x) = \frac{-13 - (9 - 13x)\ln(13)}{13^x}$

Step 2: Set $f'(x) = 0$ to find critical points

We set the numerator equal to zero to solve for $x$: $-13 - (9 - 13x)\ln(13) = 0$ Solve for $x$: $(9 - 13x)\ln(13) = -13$ $9 - 13x = \frac{-13}{\ln(13)}$ $-13x = \frac{-13}{\ln(13)} - 9$ $x = \frac{\frac{13}{\ln(13)} + 9}{13}$

This is the exact form of the critical point.

Final Answer:

The critical point is $x = \frac{\frac{13}{\ln(13)} + 9}{13}$.

Would you like more details or have any questions?

Here are some related questions to explore:

1. How do you classify a critical point as a maximum, minimum, or saddle point?
2. What role does the second derivative play in determining the nature of a critical point?
3. Can this function have multiple critical points? Why or why not?
4. What happens to the function's behavior as $x \to \infty$?
5. How does changing the base of the exponent (e.g., using 14 instead of 13) affect the critical points?

Tip: The quotient rule is essential for differentiating ratios of functions efficiently. Always simplify after applying it to avoid complicated expressions.