Question

b) $g(x) = \frac{1}{2} \cdot 4^x$
left: $\lim\limits_{x \to -\infty} f(x) =$
right: $\lim\limits_{x \to \infty} f(x) =$
increasing or decreasing
concave up or concave down
roc: increasing/decreasing

Explanation:

Step1: Analyze Left Limit ($x \to -\infty$)

For the function $g(x) = \frac{1}{2} \cdot 4^x$, recall the property of exponential functions: for $a > 1$, $\lim_{x \to -\infty} a^x = 0$. Here, $a = 4 > 1$, so $\lim_{x \to -\infty} 4^x = 0$. Then, $\lim_{x \to -\infty} \frac{1}{2} \cdot 4^x = \frac{1}{2} \cdot 0 = 0$.

Step2: Analyze Right Limit ($x \to \infty$)

For $a > 1$, $\lim_{x \to \infty} a^x = \infty$. So, $\lim_{x \to \infty} 4^x = \infty$, and thus $\lim_{x \to \infty} \frac{1}{2} \cdot 4^x = \frac{1}{2} \cdot \infty = \infty$.

Step3: Determine Increasing/Decreasing

The derivative of $g(x) = \frac{1}{2} \cdot 4^x$ is $g'(x) = \frac{1}{2} \cdot 4^x \cdot \ln(4)$. Since $\ln(4) > 0$ and $4^x > 0$ for all $x$, $g'(x) > 0$ for all $x$. So, $g(x)$ is increasing.

Step4: Determine Concavity

The second derivative: $g''(x) = \frac{1}{2} \cdot 4^x \cdot (\ln(4))^2$. Since $(\ln(4))^2 > 0$ and $4^x > 0$, $g''(x) > 0$ for all $x$. Thus, $g(x)$ is concave up.

Step5: Rate of Change (ROC)

Since $g'(x) = \frac{1}{2} \cdot 4^x \cdot \ln(4)$ and $4^x$ is increasing, $g'(x)$ is increasing (as the product of positive increasing functions). So, ROC is increasing.

Answer:

• Left Limit: $\boldsymbol{0}$
• Right Limit: $\boldsymbol{\infty}$
• Increasing or Decreasing: $\boldsymbol{\text{Increasing}}$
• Concave Up or Concave Down: $\boldsymbol{\text{Concave Up}}$
• ROC: $\boldsymbol{\text{Increasing}}$