Maximal domain and tangent of the square root function

Problem

We are given the function \(g:x\rightarrow\sqrt{3x+9}\) with maximal domain \(D\).

1. Determine \(D\) and state the zero of \(g\).

2. Determine the equation of the tangent at the graph of \(g\) at point \(P(0|3)\).

Solution of part a

By solving the condition \(\sqrt{3x+9}=0\) for \(x\), one obtains the zero at \(x=-3\).

This result can easily be checked by means of Sage:

The domain is obtained by the requirement that the argument of the square root be larger or equal zero. This is the case if \(3x+9\geq0\) or \(x\geq-3\). Thus the domain is obtained as \(D=[-3,\infty[\). The result is illustrated by the graph of the function \(g(x)\).

Solution of part b

In order to determine the equation of the tangent at the point \(P(0|3)\), we need to evaluate the derivative of \(g\) at this point. We find

\[\frac{\text{d}g}{\text{d}x} = g'(x) = \frac{3}{2\sqrt{3x+9}}.\]

As a consequence, \(g'(0)=\frac{1}{2}\).

Because of \(g(0)=3\), the tangent \(h\) at point \(P\) is given by

\[h(x) = \frac{1}{2} x +3\,.\]

This result can be graphically verified by means of Sage. The function \(g(x)\) is represented in blue while the tangent \(h(x)\) is displayed in red.