Question-35

Consider a function \(\displaystyle f:[ 0,\infty )\rightarrow \mathbb{R}\) defined as:

\[ \begin{equation*} f( x) =x+\sqrt{x} \end{equation*} \]

Which of the following is a first-order linear approximation for \(\displaystyle f\) around \(\displaystyle x=1\)?

• \(\displaystyle \frac{6x+1}{4}\)

• \(\displaystyle \frac{x+3}{2}\)

• \(\displaystyle \frac{3x+1}{2}\)

• \(\displaystyle 3x-1\)

Answer

• \(\displaystyle \frac{6x+1}{4}\)

• \(\displaystyle \frac{x+3}{2}\)

• \(\displaystyle \frac{3x+1}{2}\)

• \(\displaystyle 3x-1\)

Solution

The first order approximation is given by:

\[ \begin{equation*} L_{f}( x)\Bigl|_{x=1} =f( 1) +f^{\prime }( 1)( x-1) \end{equation*} \]

Now we compute \(f^{\prime}(x)\): \[ \begin{equation*} f^{\prime }( x) =1+\frac{1}{2\sqrt{x}} \end{equation*} \]

We see that \(\displaystyle f^{\prime }( 1) =\frac{3}{2}\), which gives the first linear approximation is:

\[ \begin{equation*} L_{f}( x)\Bigl|_{x=1} =2+\frac{3}{2}( x-1) =\frac{3x+1}{2} \end{equation*} \]