what are the features of the function $f(x) = -\\left(\\frac{1}{3}\\right)^x$ graphed below?

Explanation:

Step1: Identify domain

The function $f(x) = -\left(\frac{1}{3}\right)^x$ has no restrictions on $x$ (exponential functions accept all real numbers). Domain: $(-\infty, \infty)$

Step2: Identify range

$\left(\frac{1}{3}\right)^x > 0$ for all real $x$, so $-\left(\frac{1}{3}\right)^x < 0$. Range: $(-\infty, 0)$

Step3: Find y-intercept

Set $x=0$: $f(0) = -\left(\frac{1}{3}\right)^0 = -1$ Y-intercept: $(0, -1)$

Step4: Find x-intercept

Set $f(x)=0$: $-\left(\frac{1}{3}\right)^x = 0$ has no solution, since $\left(\frac{1}{3}\right)^x$ never equals 0. No x-intercept.

Step5: Analyze end behavior

As $x\to\infty$: $\left(\frac{1}{3}\right)^x \to 0$, so $f(x) = -\left(\frac{1}{3}\right)^x \to 0$ As $x\to-\infty$: $\left(\frac{1}{3}\right)^x = 3^{-x} \to \infty$, so $f(x) = -\left(\frac{1}{3}\right)^x \to -\infty$

Step6: Determine monotonicity

The derivative $f'(x) = -\left(\frac{1}{3}\right)^x \ln\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^x \ln(3)$ Since $\left(\frac{1}{3}\right)^x >0$ and $\ln(3)>0$, $f'(x)>0$ for all $x$. The function is strictly increasing.

Step7: Identify asymptotes

Horizontal asymptote: As $x\to\infty$, $f(x)\to0$, so $y=0$ is a horizontal asymptote. No vertical asymptotes.

Answer:

• Domain: All real numbers, $(-\infty, \infty)$
• Range: All negative real numbers, $(-\infty, 0)$
• Y-intercept: $(0, -1)$
• X-intercept: None
• End Behavior: As $x\to\infty$, $f(x)\to0$; as $x\to-\infty$, $f(x)\to-\infty$
• Monotonicity: Strictly increasing over its entire domain
• Asymptotes: Horizontal asymptote at $y=0$, no vertical asymptotes