QUESTION I 1.1 Sketch the graph of $$ f(x)=4^{\prime} $$, clearly showing ALL intercepUs and asymptote/s where necessary. Use the diagram sheet provided on (3) page 4. 1.2 Give the range of $$ f $$. (1) 1.3 Glve the equation of $$ \Gamma^{-1}(x) $$. (1) 1.4 On the same set of axes, sketch $$ f^{-1}|x| $$ clearly showing ALL intercept/s (3) and asymptote/s where nacessary. 1.5 Is $$ f^{-1}(x) $$ a one-to-one or many-to-one function? Give a reason for (1) your answer.

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**1.1 Graph of $$ f(x)=4^x $$**  
- **Intercepts:**  
  - $$y$$-intercept: When $$x=0$$,  
    $$
    f(0)=4^0=1.
    $$
    So the graph passes through $$(0,1)$$.  
  - There is no $$x$$-intercept because $$4^x>0$$ for all $$x$$.

- **Asymptote:**  
  - As $$x\to-\infty$$, $$4^x\to 0$$. Thus, the horizontal line  
    $$
    y=0 
    $$
    is a horizontal asymptote.

- **Sketch Considerations:**  
  - The function is always positive and increases as $$x$$ increases.
  - Plot a few points such as:
    - $$x=1$$: $$4^1 = 4$$
    - $$x=-1$$: $$4^{-1} = \frac{1}{4}$$
  - Draw a smooth curve passing through these points, approaching $$y=0$$ as $$x\to -\infty$$ and increasing steeply for $$x>0$$.

---

**1.2 Range of $$ f $$**  
Since $$4^x>0$$ for every real $$x$$, the range is:  
$$
(0, \infty)
$$

---

**1.3 Equation of $$ \Gamma^{-1}(x) $$**  
The inverse of $$ f(x)=4^x $$ is found by solving  
$$
y=4^x \quad \text{for } x.
$$
Taking logarithms (with base 4) on both sides gives:  
$$
x=\log_4 y.
$$
Interchanging $$x$$ and $$y$$ (to express the inverse as a function of $$x$$) results in:  
$$
\Gamma^{-1}(x)=\log_4 x.
$$

---

**1.4 Graph of $$ f^{-1}(|x|) $$**  
Here, we consider the function:  
$$
f^{-1}(|x|)=\log_4(|x|).
$$

- **Domain:**  
  - The logarithmic function $$\log_4 u$$ is defined for $$u>0$$.  
  - Since $$u=|x|$$, we require $$|x|>0$$, i.e.,  
    $$
    x\neq 0.
    $$
  
- **Intercepts and Key Points:**  
  - Solve $$\log_4(|x|)=0$$:  
    $$
    \log_4(|x|)=0 \Longrightarrow |x|=4^0=1.
    $$
    Thus, $$x=1$$ and $$x=-1$$ are the points where the graph crosses the horizontal axis (i.e., $$y=0$$).
  
- **Symmetry:**  
  - The function is even (symmetric about the $$y$$-axis) because the argument is $$|x|$$.

- **Vertical Asymptote:**  
  - As $$x$$ approaches 0, $$|x|\to 0^+$$ so $$\log_4(|x|)\to -\infty$$.  
    Hence, there is a vertical asymptote at  
    $$
    x=0.
    $$

- **Sketch Considerations:**  
  - Sketch $$\log_4(|x|)$$ for $$x>0$$ (which is just $$\log_4x$$), and then reflect it about the $$y$$-axis for $$x<0$$.
  - Clearly mark the intercepts $$(-1,0)$$ and $$(1,0)$$ and the vertical asymptote at $$x=0$$.

---

**1.5 Is $$ f^{-1}(x) $$ one-to-one or many-to-one?**  
The inverse function is  
$$
f^{-1}(x)=\log_4 x.
$$
- **Reasoning:**  
  - A function is one-to-one if each output is associated with exactly one input.  
  - The logarithmic function $$\log_4 x$$ is strictly increasing for $$x>0$$, meaning that if  
    $$
    \log_4 x_1 = \log_4 x_2,
    $$
    then  
    $$
    x_1 = x_2.
    $$
- **Conclusion:**  
  $$\log_4 x$$ is a one-to-one function.
**1.1 Graph of $$ f(x)=4^x $$:** - **Intercepts:** - $$y$$-intercept at $$(0,1)$$. - No $$x$$-intercept since $$4^x > 0$$ for all $$x$$. - **Asymptote:** - Horizontal asymptote at $$y=0$$ as $$x$$ approaches $$-\infty$$. - **Key Points:** - $$x=1$$: $$4^1 = 4$$ - $$x=-1$$: $$4^{-1} = \frac{1}{4}$$ **1.2 Range of $$ f $$:** $$(0, \infty)$$ **1.3 Equation of $$ \Gamma^{-1}(x) $$:** $$ \Gamma^{-1}(x) = \log_4 x $$ **1.4 Graph of $$ f^{-1}(|x|) $$:** - **Intercepts:** - $$(-1,0)$$ and $$(1,0)$$ - **Asymptote:** - Vertical asymptote at $$x=0$$ - **Symmetry:** - Even function symmetric about the $$y$$-axis **1.5 Is $$ f^{-1}(x) $$ one-to-one or many-to-one?** One-to-one because $$\log_4 x$$ is strictly increasing for $$x > 0$$.

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