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: Find Left Limit ($x \to -\infty$)

For the function $g(x) = \frac{1}{2} \cdot 4^x$, as $x$ approaches $-\infty$, $4^x = \frac{1}{4^{|x|}}$ (since $x$ is negative, $4^x = 4^{-|x|} = \frac{1}{4^{|x|}}$). As $|x| \to \infty$ (because $x \to -\infty$), $4^{|x|} \to \infty$, so $\frac{1}{4^{|x|}} \to 0$. Then $\frac{1}{2} \cdot 4^x \to \frac{1}{2} \cdot 0 = 0$. So $\lim_{x \to -\infty} g(x) = 0$.

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

As $x \to \infty$, $4^x$ grows without bound (since the base $4 > 1$ for exponential function $a^x$ with $a > 1$). So $\frac{1}{2} \cdot 4^x \to \infty$. Thus, $\lim_{x \to \infty} g(x) = \infty$.

Step3: Determine Increasing/Decreasing

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

Step4: Determine Concave Up/Down

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

Step5: Rate of Change (ROC)

Since the first derivative $g'(x) = \frac{1}{2} \cdot 4^x \cdot \ln(4)$ and $4^x$ is increasing (as $x$ increases, $4^x$ increases), and $\ln(4) > 0$, $\frac{1}{2} > 0$, the derivative $g'(x)$ is increasing (because $4^x$ is increasing and all other factors are positive constants). So the rate of change 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: $\boldsymbol{\text{Increasing}}$