QUESTION I I. 1 Sketch the graph of \( f(x)=4^{4} \), clearly showing All intercept/s and asymptote/s where necessary. Use the diagram sheet provided on page 4. 1.2 Glve the range of \( f \). 1.3 Glve the equation of \( f^{-1}(x) \). 1.4 On the same set of axes, sketch \( f^{-1}(x) \) clearly showing All intercept/s and asymptote/s where necessary. 1.5 Is \( f^{-1}(x) \) a one-to-one or many-to-one function? Give a reason for your answer. 1.6 Determine the values of \( x \) for which \( -1 \leq f^{-1}|x| \leq 2 \) 1.7 For which values of \( x \) will \( x . f^{-1}|x|<0 \) ?

**1.1. Sketching the graph of \( f(x)=4^x \)** - The function is exponential. - **Intercept:** \( f(0)=4^0=1 \) so the graph passes through \((0,1)\). - **Asymptote:** As \( x\to -\infty,\; 4^x\to 0 \), so there is a horizontal asymptote at \( y=0 \). - **Other properties:** The graph is increasing for all \( x \), and there is no \( x \)-intercept because \( 4^x>0 \) for all \( x \). **1.2. Range of \( f \)** - Since \( 4^x>0 \) for any real \( x \), the range is \[ (0,\infty). \] **1.3. Equation of \( f^{-1}(x) \)** - To find the inverse, write \[ y=4^x. \] - Swap \( x \) and \( y \): \[ x=4^y. \] - Solve for \( y \) by taking logarithms with base 4: \[ y=\log_{4}(x). \] - Thus, \[ f^{-1}(x)=\log_{4}(x). \] **1.4. Sketching the graph of \( f^{-1}(x)=\log_{4}(x) \)** - **Intercept:** \( f^{-1}(1)=\log_4(1)=0 \) so the graph passes through \((1,0)\). - **Asymptote:** As \( x\to 0^+,\; \log_{4}(x)\to -\infty\), so there is a vertical asymptote at \( x=0 \). - **Other properties:** The graph is increasing and is the reflection of \( f(x)=4^x \) about the line \( y=x \). **1.5. Nature of \( f^{-1}(x) \)** - \( f^{-1}(x)=\log_{4}(x) \) is a one-to-one function because for each \( x>0 \) there is a unique output, and it passes the horizontal line test. **1.6. Values of \( x \) for which \(-1 \leq f^{-1}(|x|) \leq 2\)** - Here, \( f^{-1}(|x|)=\log_{4}(|x|) \). We need to solve \[ -1\leq \log_{4}(|x|) \leq 2. \] _Step 1: Solve \(\log_{4}(|x|)\geq -1\)_ - Rewrite in exponential form: \[ |x|\geq 4^{-1}=\frac{1}{4}. \] _Step 2: Solve \(\log_{4}(|x|)\leq 2\)_ - Rewrite in exponential form: \[ |x|\leq 4^{2}=16. \] - Combining these, we have \[ \frac{1}{4}\leq |x|\leq 16. \] - In interval notation, this is \[ x\in \left[-16,-\frac{1}{4}\right] \cup \left[\frac{1}{4},16\right]. \] **1.7. Values of \( x \) for which \( x\cdot f^{-1}(|x|)<0 \)** - Recall that \( f^{-1}(|x|)=\log_{4}(|x|) \). We therefore need \[ x\cdot\log_{4}(|x|)<0. \] Consider two cases based on the sign of \( x \): **Case 1: \( x>0 \)** - For \( x>0 \), \( |x|=x \) and the inequality becomes \[ x\log_{4}(x)<0. \] - Since \( x>0 \), the sign of the product depends on \(\log_{4}(x)\). \(\log_{4}(x)<0\) when \( 00 \), the inequality holds when \[ 00\) when \( |x|>1 \) (i.e. when the absolute value of \( x \) is greater than 1). Since \( x<0 \), this translates to \[ x<-1. \] - Combining both cases, the inequality \( x\cdot f^{-1}(|x|)<0 \) holds when \[ x<-1 \quad \text{or} \quad 0