(c) Given $$ f(x)=-\sqrt{x} $$. (1) Write down the domain and range of $$ f $$. (2) Write down the equation of $$ f^{-1} $$. (3) Sketch the graphs of $$ f $$ and $$ f^{-1} $$ on the same set of axes. (4) Determine the equation of (i) $$ \quad g $$, the reflection of $$ f $$ in the $$ x $$-axis. (ii) $$ \quad h $$, the reflection of $$ f $$ in the $$ y $$-axis. (d) The sketch alongside shows the graph of $$ f(x)=a x^{2} $$, with a restriction on its domain, and the graph of $$ f^{-1} $$. $$ f $$ passes through the point $$ (\sqrt{2} ; 4) $$. P is the point of intersection of $$ f $$ and $$ f^{-1} $$. (1) Write down the domain of $$ f $$. (2) Determine the value of $$ a $$. (3) Calculate the coordinates of P . (4) For which values of $$ x $$ is $$ f(x)f^{-1}(x) $$ ?

Question

(c) Given $$ f(x)=-\sqrt{x} $$. (1) Write down the domain and range of $$ f $$. (2) Write down the equation of $$ f^{-1} $$. (3) Sketch the graphs of $$ f $$ and $$ f^{-1} $$ on the same set of axes. (4) Determine the equation of (i) $$ \quad g $$, the reflection of $$ f $$ in the $$ x $$-axis. (ii) $$ \quad h $$, the reflection of $$ f $$ in the $$ y $$-axis. (d) The sketch alongside shows the graph of $$ f(x)=a x^{2} $$, with a restriction on its domain, and the graph of $$ f^{-1} $$. $$ f $$ passes through the point $$ (\sqrt{2} ; 4) $$. P is the point of intersection of $$ f $$ and $$ f^{-1} $$. (1) Write down the domain of $$ f $$. (2) Determine the value of $$ a $$. (3) Calculate the coordinates of P . (4) For which values of $$ x $$ is $$ f(x)>f^{-1}(x) $$ ?

UpStudy AI · Accepted Answer

Alright, I need to tackle this multi-part problem involving functions, their inverses, reflections, and domain/range determinations. Let's break it down step by step to ensure I cover all aspects thoroughly.

**Part (c):**

1. **Domain and Range of $$ f(x) = -\sqrt{x} $$:**
   - The square root function $$ \sqrt{x} $$ is defined for $$ x \geq 0 $$. Therefore, the domain of $$ f(x) $$ is $$ x \geq 0 $$.
   - The range of $$ \sqrt{x} $$ is $$ y \geq 0 $$. Since $$ f(x) = -\sqrt{x} $$, the range becomes $$ y \leq 0 $$.

2. **Inverse Function $$ f^{-1} $$:**
   - To find the inverse, I'll swap $$ x $$ and $$ y $$ and solve for $$ y $$:
     $$
     x = -\sqrt{y} \implies \sqrt{y} = -x \implies y = x^2
     $$
   - However, since $$ \sqrt{y} $$ is always non-negative, $$ x $$ must be non-positive. Therefore, the domain of $$ f^{-1} $$ is $$ x \leq 0 $$, and the range is $$ y \geq 0 $$.

3. **Graphs of $$ f $$ and $$ f^{-1} $$:**
   - $$ f(x) = -\sqrt{x} $$ is a reflection of $$ \sqrt{x} $$ across the x-axis.
   - $$ f^{-1}(x) = x^2 $$ is a parabola opening upwards.
   - Both graphs will intersect at the origin (0,0).

4. **Reflections:**
   - **(i) Reflection in the x-axis ($$ g $$):**
     - Reflecting $$ f(x) $$ across the x-axis changes the sign of the output:
       $$
       g(x) = -f(x) = \sqrt{x}
       $$
   - **(ii) Reflection in the y-axis ($$ h $$):**
     - Reflecting $$ f(x) $$ across the y-axis changes the sign of the input:
       $$
       h(x) = f(-x) = -\sqrt{-x}
       $$
     - However, $$ \sqrt{-x} $$ is only defined for $$ x \leq 0 $$, so the domain of $$ h $$ is $$ x \leq 0 $$.

**Part (d):**

1. **Domain of $$ f(x) = a x^{2} $$:**
   - The domain of a quadratic function is all real numbers, $$ x \in \mathbb{R} $$.

2. **Determining the Value of $$ a $$:**
   - Given that $$ f $$ passes through the point $$ (\sqrt{2}, 4) $$:
     $$
     4 = a (\sqrt{2})^2 \implies 4 = a \times 2 \implies a = 2
     $$
   - Therefore, $$ f(x) = 2x^2 $$.

3. **Coordinates of Point $$ P $$:**
   - Point $$ P $$ is the intersection of $$ f(x) $$ and $$ f^{-1}(x) $$.
   - Since $$ f(x) = 2x^2 $$, its inverse $$ f^{-1}(x) = \sqrt{\frac{x}{2}} $$.
   - Setting $$ f(x) = f^{-1}(x) $$:
     $$
     2x^2 = \sqrt{\frac{x}{2}} \implies 4x^4 = \frac{x}{2} \implies 8x^4 = x \implies x(8x^3 - 1) = 0
     $$
   - Solutions: $$ x = 0 $$ or $$ x = \frac{1}{2} $$.
   - For $$ x = 0 $$:
     $$
     f(0) = 0 \quad \text{and} \quad f^{-1}(0) = 0
     $$
   - For $$ x = \frac{1}{2} $$:
     $$
     f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 = \frac{1}{2} \quad \text{and} \quad f^{-1}\left(\frac{1}{2}\right) = \sqrt{\frac{\frac{1}{2}}{2}} = \frac{1}{2}
     $$
   - Therefore, the points of intersection are $$ (0, 0) $$ and $$ \left(\frac{1}{2}, \frac{1}{2}\right) $$.

4. **Values of $$ x $$ where $$ f(x) > f^{-1}(x) $$:**
   - From the intersection points, $$ f(x) > f^{-1}(x) $$ when $$ x > \frac{1}{2} $$.
   - Therefore, $$ f(x) > f^{-1}(x) $$ for $$ x > \frac{1}{2} $$.

**Part (c):** 1. **Domain and Range of $$ f(x) = -\sqrt{x} $$:** - *Domain:* $$ x \geq 0 $$ - *Range:* $$ y \leq 0 $$ 2. **Inverse Function $$ f^{-1} $$:** - $$ f^{-1}(x) = x^2 $$ with domain $$ x \leq 0 $$ and range $$ y \geq 0 $$ 3. **Graphs of $$ f $$ and $$ f^{-1} $$:** - $$ f(x) = -\sqrt{x} $$ is a reflection of $$ \sqrt{x} $$ across the x-axis. - $$ f^{-1}(x) = x^2 $$ is a parabola opening upwards. - Both graphs intersect at the origin (0,0). 4. **Reflections:** - **(i) Reflection in the x-axis ($$ g $$):** $$ g(x) = \sqrt{x} $$ - **(ii) Reflection in the y-axis ($$ h $$):** $$ h(x) = -\sqrt{-x} $$ with domain $$ x \leq 0 $$ **Part (d):** 1. **Domain of $$ f(x) = a x^{2} $$:** - All real numbers, $$ x \in \mathbb{R} $$ 2. **Determining the Value of $$ a $$:** - Given $$ f(\sqrt{2}) = 4 $$: $$ 4 = a (\sqrt{2})^2 \implies a = 2 $$ - Therefore, $$ f(x) = 2x^2 $$ 3. **Coordinates of Point $$ P $$:** - Intersection points are $$ (0, 0) $$ and $$ \left(\frac{1}{2}, \frac{1}{2}\right) $$ 4. **Values of $$ x $$ where $$ f(x) > f^{-1}(x) $$:** - $$ f(x) > f^{-1}(x) $$ for $$ x > \frac{1}{2} $$

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