(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 \( f(x)>f^{-1}(x) \) ?

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 \( x > \frac{1}{2} \).