Determine \( f^{-1}(x) \) from first principles if \( f(x)=3 x^{2} \) John determines \( g^{\prime}(a) \), the derivative of a certain function \( g \) at \( x=a \), and arrives at the answer. \( \lim _{x \rightarrow 0} \frac{\sqrt{4+h}-2}{h} \) Write down the equation of \( g \) and the value of \( a \). Determine \( \frac{d y}{d x} \) if \( y=\sqrt{x^{3}}-\frac{5}{x^{3}} \) \( g(x)=-8 x+20 \) is a tangent to \( \quad f(x)=x^{3}+a x^{2}+b x+18 \) at \( x=1 \). Calculate the values of \( a \) and \( b \).

To find \( f^{-1}(x) \) where \( f(x) = 3x^2 \), we start with \( y = 3x^2 \). Rearranging for \( x \) gives \( x^2 = \frac{y}{3} \), so \( x = \sqrt{\frac{y}{3}} \). Thus, \( f^{-1}(x) = \sqrt{\frac{x}{3}} \). This inverse is only valid for \( x \geq 0 \) since \( f(x) = 3x^2 \) is a non-negative function. When John computes \( g^{\prime}(a) \) from \( \lim_{h \rightarrow 0} \frac{\sqrt{4+h}-2}{h} \), he recognizes this limit as the derivative of \( g(x) = \sqrt{x} \) evaluated at \( x = 4 \). Therefore, the value of \( a \) is \( 4 \). To determine \( \frac{dy}{dx} \) for \( y = \sqrt{x^3} - \frac{5}{x^3} \), we apply the power rule. This gives \( \frac{dy}{dx} = \frac{3}{2}x^{1/2} - \left(-\frac{15}{x^4}\right) = \frac{3}{2\sqrt{x}} + \frac{15}{x^4} \). Given that \( g(x) = -8x + 20 \) is a tangent to \( f(x) = x^3 + ax^2 + bx + 18 \) at \( x = 1 \), we set \( f(1) = g(1) \). Equating \( f(1) = 1 + a + b + 18 \) to \( 12 \) gives us \( a + b = -7 \). The equality of derivatives yields \( 3 + 2a + b = -8 \). Solving these equations, we find \( a = -5 \) and \( b = -2 \).