be a furnction. 4 Given: $$ f(x)=\frac{1}{2} x^{2} $$ and $$ g(x)=2 x+4 $$ 4.1 Is $$ f(x) $$ a function? Give a reason for your answer. 4.2 Write down the domain of $$ f(x) $$ for $$ f^{-1}(x) $$ to be a function 4.3 Write down the equation of $$ g^{-1}(x) $$ in the form $$ y=\ldots $$ 4.4 Write down the range of $$ f(x) $$.

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Certainly! Let's address each part of the problem step by step.

### **Given:**
- $$ f(x) = \frac{1}{2} x^{2} $$
- $$ g(x) = 2x + 4 $$

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### **4.1 Is $$ f(x) $$ a function?**

**Answer:** Yes, $$ f(x) = \frac{1}{2} x^{2} $$ is a function.

**Reason:** For every input value $$ x $$ in the domain, there is exactly one output value $$ f(x) $$. This satisfies the definition of a function, where each input is associated with a single output.

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### **4.2 Domain of $$ f(x) $$ for $$ f^{-1}(x) $$ to be a function**

**Answer:** The domain of $$ f(x) $$ must be restricted to $$ x \geq 0 $$ (i.e., all non-negative real numbers).

**Explanation:** The function $$ f(x) = \frac{1}{2} x^{2} $$ is not one-to-one over all real numbers because both $$ x $$ and $$ -x $$ yield the same output. To ensure that the inverse $$ f^{-1}(x) $$ is also a function (i.e., passes the horizontal line test), we need to restrict the domain of $$ f(x) $$ to either $$ x \geq 0 $$ or $$ x \leq 0 $$. Typically, the domain is restricted to $$ x \geq 0 $$.

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### **4.3 Equation of $$ g^{-1}(x) $$**

**Answer:** The inverse function $$ g^{-1}(x) $$ is given by:
$$
y = \frac{x - 4}{2}
$$

**Derivation:**
1. Start with $$ y = 2x + 4 $$.
2. Swap $$ x $$ and $$ y $$: $$ x = 2y + 4 $$.
3. Solve for $$ y $$:
   $$
   x - 4 = 2y \
   y = \frac{x - 4}{2}
   $$
   So, $$ g^{-1}(x) = \frac{x - 4}{2} $$.

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### **4.4 Range of $$ f(x) $$**

**Answer:** The range of $$ f(x) = \frac{1}{2} x^{2} $$ is all real numbers $$ y $$ such that $$ y \geq 0 $$.

**Explanation:** Since $$ f(x) $$ is a quadratic function that opens upwards and its minimum value occurs at $$ x = 0 $$, where $$ f(0) = 0 $$, the function outputs are all non-negative real numbers. Therefore, the range is $$ y \geq 0 $$.

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If you have any further questions or need more clarification on any part, feel free to ask!
**4.1** Yes, $$ f(x) = \frac{1}{2} x^{2} $$ is a function because each input $$ x $$ has exactly one output. **4.2** To make $$ f^{-1}(x) $$ a function, the domain of $$ f(x) $$ should be restricted to $$ x \geq 0 $$. **4.3** The inverse function $$ g^{-1}(x) $$ is $$ y = \frac{x - 4}{2} $$. **4.4** The range of $$ f(x) $$ is all real numbers $$ y $$ where $$ y \geq 0 $$.

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