Question-31

Suppose \(\displaystyle f\) and \(\displaystyle g\) are functions on \(\displaystyle \mathbb{R}\) such that:

\[ \begin{equation*} \begin{aligned} f( 0) & =4\\ f^{\prime }( 0) & =3\\ f^{\prime }( 3) & =-1 \end{aligned} \ \ \ \ \ \ \ \ \begin{aligned} g( 0) & =3\\ g^{\prime }( 0) & =-2 \end{aligned} \end{equation*} \]

Which of the following are true?

• \(\displaystyle ( f\circ g)^{\prime }( 0) =2\)

• \(\displaystyle ( fg)^{\prime }( 0) =-6\)

• \(\displaystyle \left(\frac{f}{g}\right)( 0) =\frac{-17}{9}\)

• \(\displaystyle \left(\frac{f}{g}\right)( 0) =\frac{17}{9}\)

Hint

• Chain rule
• Product rule
• Quotient rule

Answer

• \(\displaystyle ( f\circ g)^{\prime }( 0) =2\)

• \(\displaystyle ( fg)^{\prime }( 0) =-6\)

• \(\displaystyle \left(\frac{f}{g}\right)( 0) =\frac{-17}{9}\)

• \(\displaystyle \left(\frac{f}{g}\right)( 0) =\frac{17}{9}\)

Solution

Chain rule

Using the chain rule, we have:

\[ \begin{equation*} ( f\circ g)( x) =f( g( x)) \Longrightarrow ( f\circ g)^{\prime }( x) =f^{\prime }( g( x)) g^{\prime }( x) \end{equation*} \]

For \(\displaystyle x=0\), we get \(\displaystyle f^{\prime }( g( 0)) g^{\prime }( 0) =f^{\prime }( 3) g^{\prime }( 0) =( -1)( -2) =2\)

Product rule

Next, using the product rule:

\[ \begin{equation*} ( fg)( x) =f( x) g( x) \Longrightarrow ( fg)^{\prime }( x) =f^{\prime }( x) g( x) +f( x) g^{\prime }( x) \end{equation*} \]

For \(\displaystyle x=0\), we get this to be \(\displaystyle 3\times 3+4\times ( -2) =1\)

Quotient rule

Finally, using the quotient rule:

\[ \begin{equation*} \left(\frac{f}{g}\right)( x) =\frac{f( x)}{g( x)} \Longrightarrow \left(\frac{f}{g}\right)^{\prime }( x) =\frac{g( x) f^{\prime }( x) -f( x) g^{\prime }( x)}{[ g( x)]^{2}} \end{equation*} \]

For \(\displaystyle x=0\), we get \(\displaystyle \frac{3\times 3-4\times ( -2)}{3^{2}} =\frac{17}{9}\).

Note that the function \(\cfrac{f}{g}(x)\) is defined only for those points where \(g(x) \neq 0\). Since \(g(0) \neq 0\), the function is defined at \(x = 0\).